ANALISIS MATEMATICO 2013.docx

UNIVERSIDAD NACIONAL DE HUANCAVELICA

FACULTAD DE CIENCIAS DE INGENIERÍA

ESCUELA ACADÉMICO PROFESIONAL DE INGENIERÍA CIVIL-HUANCAVELICA

I.- DERIVADAS1.- encontrar la derivada de las funciones siguientes, usando solamente la definición de derivada:

a) f ( x )=√x2+3 x−e2x (yendi)solucion

f ' ( x )=límh→0

(√( x+h )2+3 ( x+h )−e2 ( x+h )−√x2+3 x−e2x )h

f ' ( x )=límh→0

(√( x+h )2+3 ( x+h )−e2 ( x+h )−√x2+3 x−e2x ) (√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x )h(√ ( x+h )2+3 (x+h )−e2 ( x+h)+√ x2+3 x−e2 x)

f ' ( x )= límh→ 0

(√( x+h )2+3 ( x+h )−e2 ( x+h ))2−(√x2+3 x−e2x )2

h (√( x+h )2+3 ( x+h )−e2( x+h)+√x2+3 x−e2x )

f ' ( x )=límh→0

(( x+h )2+3 ( x+h )−e2 ( x+h )−x2−3x+e2x )h (√ ( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x )

f ' ( x )=límh→0

(x2+h2+2xh+3x+3h−e2 (x +h)−x2−3 x+e2x )h (√ ( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3x−e2x)

ANALISIS MATEMÁTICO III




f ' ( x )=límh→0

(h2+2xh+3h−e2 ( x+h )+e2 x)h (√ ( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x )

f ' ( x )=límh→0

h2+2 xh+3h

h (√ ( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x )+ lím

h→0

e2x−e2 ( x+h )

(√ (x+h )2+3 ( x+h )−e2 (x+h)+√ x2+3 x−e2x )

f ' ( x )=límh→0

h+2 x+3

(√( x+h )2+3 ( x+h )−e2 ( x+h )+√x2+3 x−e2x )+ lím

h→0

e2x−e2 ( x+h )

(√ (x+h )2+3 ( x+h )−e2 (x +h)+√ x2+3 x−e2x )Reemplazamos el valor de h=0

f ' ( x )= 0+2 x+3

(√ ( x+0 )2+3 ( x+0 )−e2 ( x+0 )+√x2+3 x−e2x )+ e2x−e2 ( x+0 )

(√ (x+0 )2+3 ( x+0 )−e2( x+0)+√ x2+3 x−e2x )

f ' ( x )= 2 x+32√x2+3 x−e2x

+ 2e2x

2√x2+3 x−e2x

f ' ( x )= 2x+3+2e2x

2√x2+3 x−e2x

b) g ( x )=3√ x3−2x2+x (carlos)solucion





g ' ( x )=límh→0

3√( x+h )3−2 (x+h )2+ (x+h )−3√ x3−2x2+xh

Como el índice de los radicales es 3; entonces aplicamos el criterio de factor Racionalizarte. al numerador por lo cual.

( 3√( x+h )3−2 (x+h )2+ (x+h ) )2+ 3√ ( x+h )3−2 ( x+h )2+ ( x+h ) . 3√ x3−2 x2+x+( 3√x3−2 x2+x )2=F . R .

Es decir:

g ' ( x )=límh→0

( 3√ ( x+h )3−2 ( x+h )2+( x+h )− 3√x3−2 x2+x )F .R

h(F .R)

g ' ( x )=límh→0

( 3√ ( x+h )3−2 ( x+h )2+( x+h ) )3−( 3√ x3−2 x2+x )3

h(F . R)

g ' ( x )=límh→0

x3+3 x2h+3x h2+h3−2x2−2h2−4 xh+x+h−x3+2x2−xh (F . R)

g ' ( x )=límh→0

3x2h+3 xh2+h3−2h2−4 xh+hh (F .R )

g ' ( x )=límh→0

h (3 x2+3 xh+h2−2h2−4 x+1 )

h( 3√ ( x+h )3−2 ( x+h )2+( x+h ) )2+ 3√( x+h )3−2 ( x+h )2+( x+h ) . 3√x3−2x2+x+( 3√ x3−2x2+x )2

Reemplazando cuando h=0





g ' ( x )=(3 x2+3 x (0)+(0)2−2(0)2−4 x+1 )

( 3√ ( x+0 )3−2 ( x+0 )2+( x+0 ) )2+ 3√( x+0 )3−2 ( x+0 )2+( x+0 ) . 3√x3−2 x2+x+( 3√ x3−2x2+x )2

g ' ( x )=(3 x2−4 x+1 )

( 3√ x3−2x2+x )2+ 3√ x

3−2x2+x .3√ x

3−2 x2+x+( 3√x

3−2 x2+x )2

g ' ( x )=(3x2−4 x+1 )

3√(x3−2x2+x )2+3√(x3−2x2+x )

2+3√( x3−2x2+x )

2

g ' ( x )=(3 x2−4 x+1 )

33√(x3−2 x2+ x )

2

c) h ( x )=5√2 x3−3x2+2 (heber)solucion

h ' ( x )=límh→0

5√2 ( x+h )3−3 (x+h )2+2−5√2x3−3 x2+2h

Como el índice de los radicales es 5;

Hacemos: A=2 ( x+h )3−3 ( x+h )2+2 y B=2 x3−3 x2+2





→5√A4+ 5√A3 . 5√B+ 5√A2 .

5√B2+ 5√A .5√B3+ 5√B4=F . R

Es decir:

( 5√A−5√B ) ( F . R )=( A−B )

h' (x )=límh→0

2 ( x+h )3−3 (x+h )2+2−2 x3+3 x2−2h ( F . R )

h' (x )=límh→0

2x3+6 x2h+6 xh2+2h3−3x2−3h2−6 xh+2−2 x3+3 x2−2h ( F . R )

h' (x )=límh→0

6 x2h+6 x h2+2h3−3h2−6 xhh (F .R )

h' (x )=límh→0

h (6 x2+6 xh+2h2−3h−6x )h ( F . R )

=límh→0

(6 x2+6 xh+2h2−3h−6 x )5√A4+

5√A3.5√B+

5√A2 .5√B2+ 5√A .

5√B3+5√B4

Reemplazando el valor de h=o

h' (x )=(6 x2+6 x (0)+2 (0 )2−3(0)−6 x )

5√A4+5√A3 .

5√B+5√A2 .

5√B2+ 5√A .5√B3+

5√B4

h' (x )=(6 x2−6 x )

5√A4+5√A3 .

5√B+5√A2 .

5√B2+ 5√A .5√B3+

5√B4





A=2 ( x+h )3−3 ( x+h )2+2A=2x3−3 x2+2=BReemplazando los valores

h' (x )=(6x2−6x )

5√(2 x3−3 x2+2 )4+ 5√ (2 x3−3 x2+2 )4+ 5√ (2 x3−3x2+2 )4+ 5√(2x3−3 x2+2 )4+ 5√(2 x3−3 x2+2 )4

h' (x )=(6 x2−6 x )

55√(2 x3−3 x2+2 )4

d) i (x )=3√x4+3x−√x2+9 x (carlos)solucion

i ' (x )=límh→o

3√( x+h )4+3 (x+h )−√ ( x+h )2+9 ( x+h )− 3√x4+3 x−√ x2+9 xh

i ' (x )=límh→o

[ ( 3√ (x+h )4+3 ( x+h )− 3√x4+3 x )h

−(√( x+h )2+9 ( x+h )−√x2+9 x )

h ]i ' (x )=lím

h→o

( 3√( x+h )4+3 ( x+h )−3√ x4+3 x )h

−límh→0

(√ ( x+h )2+9 ( x+h )−√x2+9 x )h





A B

Resolviendo “A”

límh→o

( 3√ ( x+h )4+3 ( x+h )− 3√x4+3 x )h

FR=3√( x+h )4+3 (x+h )2+ 3√( x+h )4+3 ( x+h ) . 3√ x4+3 x+

3√x4+3 x2

límh→o

( 3√ ( x+h )4+3 ( x+h )− 3√x4+3 x ) ( 3√ (x+h )4+3 ( x+h )2+ 3√ (x+h )4+3 ( x+h ) . 3√x4+3 x+3√ x4+3 x2 )

h(F . R)

límh→o

3√( (x+h )4+3 ( x+h ) )3−3√(x 4+3x )3

h(F .R)=lím

h→o

( x+h )4+3 ( x+h )−x4−3 xh (F .R)

límh→o

x4+h4+6 x2h2+4 x3h+4 xh3+3 x+3h−x4−3 xh(F . R)

límh→o

h4+6 x2h2+4 x3h+4 xh3+3hh(F . R)

límh→o

h (h3+6 x2h+4 x3+4 x h2+3 )h(

3√ ( x+h )4+3 ( x+h )2+ 3√ ( x+h )4+3 ( x+h ) . 3√x4+3x+3√ x4+3 x2)





límh→o

(h3+6 x2h+4 x3+4 x h2+3 )(3√ ( x+h )4+3 ( x+h )2+ 3√ ( x+h )4+3 ( x+h ) . 3√x4+3x+

3√ x4+3 x2)


¿((0)3+6 x2(0)+4 x3+4 x (0)2+3 )

(3√ ( x+0 )4+3 ( x+0 )2+ 3√ ( x+0 )4+3 ( x+0 ) . 3√ x4+3 x+

3√x4+3 x2)

A=(4 x3+3 )3√(x4+3 x )2

Resolviendo “B”

límh→0

(√ ( x+h )2+9 ( x+h )−√x2+9 x )h

límh→ 0

(√( ( x+h )2+9 ( x+h ) )2−√ (x2+9 x )2)h(√ ( x+h )2+9 ( x+h )+√ x2+9 x )

límh→0

( ( x+h )2+9 (x+h )−(x2+9 x ))h (√ ( x+h )2+9 ( x+h )+√ x2+9x )

límh→0

x2+h2+2xh+9x+9h−x2−9 x

h (√ ( x+h )2+9 ( x+h )+√ x2+9x )





límh→ 0

h (h+2 x+9 )

h (√ ( x+h )2+9 ( x+h )+√ x2+9x )Reemplazando cuando h=0

¿(0+2x+9 )

(√ ( x+0 )2+9 ( x+0 )+√ x2+9 x )

B=(2x+9 )

2 (√ x2+9 x )

i' ( x )=A−B

i' ( x )=(4 x3+3 )3√ (x4+3 x )2

−(2 x+9 )

2 (√ x2+9x )

e) k ( x )=√cos (3 x) (carlos)solucion

k ' (x)=límh→0

√cos3 ( x+h )−√cos (3x )h





k ' (x)=límh→0

√cos3 ( x+h )−√cos (3x )(√cos 3 ( x+h )+√cos (3x ))h(√cos 3 (x+h )+√cos (3x ))

k ' (x)=límh→0

(√cos3 ( x+h ))2−(√cos (3 x ) )2

h (√cos3 ( x+h )+√cos (3 x ) )

k ' ( x )=límh→0

cos3 ( x+h )−cos (3x )h (√cos3 ( x+h )+√cos (3 x ) )

=límh→0

cos3 x .cos3h−sen 3x . sen3h−cos (3 x )h (√cos3 ( x+h )+√cos (3 x ) )

k ' ( x )=límh→0

cos3 x (cos 3h−1 )−sen 3x . sen3h

h (√cos3 (x+h )+√cos (3 x ))

k ' ( x )=límh→0

cos3 x (cos 3h−1 )h (√cos3 ( x+h )+√cos (3 x ) )

−límh→0

sen3x . sen3h

h (√cos3 ( x+h )+√cos (3 x ) )

k ' ( x )=−cos3 x límh→0

(1−cos3h)

h (√cos 3 (x+h )+√cos (3 x ) )−sen3 x ( límh→ 0

sen 3h3h )( límh→ 0

3

h (√cos3 ( x+h )+√cos (3 x ) ) )k ' ( x )=−cos3 x lím

h→0

(1−cos23h)

h (√cos 3 (x+h )+√cos (3 x ) ) (1+cos3h)−sen 3x ( límh→ 0

sen3h3h )( límh→ 0

3

h (√cos3 ( x+h )+√cos (3x ) ) )

1





k ' ( x )=−cos3 x límh→0

sen23h

h (√cos 3 (x+h )+√cos (3 x ) ) (1+cos3h)−¿ sen 3x ( límh→0

3

h (√cos3 ( x+h )+√cos (3 x ) ) )¿k ' ( x )=−cos3 x lím

h→0( sen 3h3h ) límh→0

3 sen3h

h (√cos 3 (x+h )+√cos (3 x ) ) (1+cos3h)−¿ s en3 x( límh→0

3

h (√cos 3 (x+h )+√cos (3x ) ) )¿Reemplazando cuando h=0

k ' ( x )=−cos3 x (1 ) (0 )−sen3x (1)( 3

√cos3 x+√cos (3x ) )k ' ( x )=0− 3 sen3 x

2√cos3 x

k ' ( x )=−3 sen3 x

2√cos3 x

f) l (x )=√sen ( x ) (josue)solucion

l '(x )=límh→0

√sen ( x+h )−√sen ( x )h

l '(x )=límh→0

√sen ( x+h )−√sen ( x )(√sen ( x+h )+√sen ( x ))h(√sen ( x+h )+√sen (x ))





l '(x )=límh→0

(√sen ( x+h ))2−(√ sen ( x ))2

h (√sen ( x+h )+√sen ( x ) )

k ' ( x )=límh→0

sen ( x+h )−sen ( x )h (√sen ( x+h )+√sen ( x ) )

=límh→0

sen x .cosh+cos x . sen ( h )−cos xh (√cos (x+h )+√cos x )

k ' ( x )=límh→0

senx (cos h−1 )+cos x . sen (h )h (√sen (x+h )+√sen x )

k ' ( x )=límh→0

senx (cosh−1 )h (√sen ( x+h )+√sen x )

+ límh→0

cos x . sen ( h )h (√sen ( x+h )+√sen x )

k ' ( x )=(−sen x ) límh→0

(1−cos (h)) (1+cos(h))h (√sen ( x+h )+√sen x ) (1+cos (h))

+ (cos x ) límh→0

sen (h )h (√sen ( x+h )+√sen x )

k ' ( x )=(−sen x ) límh→ 0

(1−cos2(h))h (√sen ( x+h )+√sen x ) (1+cos (h))

+ (cos x )∗( lím→ 0

sen (h )h )∗(límh→0

1

h (√sen ( x+h )+√sen x ) )k ' ( x )=(−sen x ) lím

h→0

sen2h

h (√sen ( x+h )+√sen x ) (1+cos (h))+ (cos x )∗( lím→0

sen (h )h )∗(límh→0

1

h (√sen ( x+h )+√sen x ) )





k ' ( x )=(−sen x )(límh→0

senhh )(límh→0

senh

(√sen ( x+h )+√sen x ) (1+cos (h ) ) )+(cos x )(lím→0 sen (h )h )( límh→0

1

(√sen ( x+h )+√sen x) )


k ' ( x )=(−sen x ) (1 ) (0 )+(cos x )(1)( 1

√sen x+√sen x )k ' ( x )=(0)+( (cos x )

2√sen x )k ' ( x )= (cos x )

2√ sen x

g) m ( x )=tan3( 2 x3 ) (josue)

soluc ion

tan3( 2x3 )=sen3( 2 x

3 )cos3( 2x3 )





m' (x)=límh→0

[ sen3( 2 (x+h )3 )

cos3( 2 ( x+h )3 )

−sen3( 2 x

3 )cos3( 2 x

3 ) ]h

=límh→0

sen3( 2 x+2h3 )∗cos3( 2x3 )−sen3( 2x3 )∗cos3( 2 x+2h

3 )h[cos3( 2 x+2h

3 )cos3( 2 x3 )]

m' (x)=límh→0

(sen ( 2x3 )cos ( 2h3 )+cos (2h3 )sen( 2 x3 ))

3

∗cos3(2 x3 )−sen3( 2 x

3 )∗(cos ( 2x3 )cos ( 2h3 )−sen( 2h3 )sen ( 2x3 ))3

h[cos3( 2 x+2h3 )cos3( 2 x

3 )]m' (x)=lím

h→0(sen3( 2 x3 )cos3( 2h3 )+cos3( 2h3 )sen3( 2 x

3 )+3 sen2( 2x3 )cos (2h3 )+3 sen (2 x3 )cos2( 2h3 ))∗cos3( 2 x

3 )

tan3 (2 x3 )=2∗33 tan2( 2x3 )=2 tan2( 2x3 )∗sec 2( 2 x

3 )Hacemos:

a=2x3

, b=2h3

, Sa=sen (a ) , Sb=sen (b ) ,Ca=cos (a ) y Cb=cos (b)

m' (x)=lím

h→ 0(Sa3Cb3+Ca3Sb3+3 Sa2Cb+3 SaCb2 )∗Ca3−Sa3 (Ca3Cb3−Sa3Sb3−3Ca2Cb2∗Sa .Sb+3Ca .Cb∗Sa2Sb2 )





m' (x)=límh→0

(Sa3Cb3Ca3+Ca6Sb3+3Sa2CbCa3+3SaCb2Ca3 )−Ca3Cb3 Sa3+Sb3Sa6+3Ca2Cb2Sa4Sb−3Ca .CbSa5Sb2

m' (x)=límh→0

Sb3 (Ca6+Sa6 )+3Sb2 .Cb (SaCa5−CaSa5 )+3Cb2Sb (Sa2Ca4+Sa4Ca2 )h (C (a+b))3Ca3

m' (x)=límh→0

Sb3 (Ca6+Sa6 )h (C (a+b))3Ca3

+ límh→0

3 Sb2 .Cb (SaCa5−Ca Sa5 )h (C(a+b))3Ca3

+ límh→0

3Cb2Sb (Sa2Ca4+Sa4Ca2 )h (C (a+b))3Ca3

m' (x)=( límh→0

Sb2h3 )( límh→0

2Sb2 (Ca6+Sa6 )(C (a+b ) )3Ca3 )+(límh→0

Sb2h3 )( límh→0

2Sb .Cb (SaCa5−CaSa5)(C (a+b ) )3Ca3 )

+(límh→0

Sb2h3 )( límh→0

2Cb2 (Sa2Ca4+Sa4Ca2 )(C (a+b))3Ca3 )

m' (x)= (1 )∗(0 )+(1 ) (0 )+(1 )∗( límh→0

2Cb2Sa2Ca2 (Ca2+Sa2 )(C (a+b))3Ca3 )

m' ( x )=( límh→02)∗lím

h→0

Cb2Sa2

(C(a+b))3Ca





m' ( x )=2∗límh→0

Cb2Sa2

(C(a+b))3Ca


m' ( x )=2∗límh→0

cos ( 2h3 )2

sen( 2 x3 )

2

cos( 2 ( x+h )3 )

3

cos (2 x3 )

=¿2∗1∗sen (2 x

3 )2

cos( 2 ( x+0 )3 )

3

cos( 2 x3 )

¿

m' ( x )=2 sen ( 2x3 )

2

cos( 2 x3 )

4 =2 tan2 2x3

∗sec22x3

h) ñ ( x )=log(√ x) (josue)solucion

ñ '( x)=límh→0

log (√ x+h )−log(√ x)h

=límh→0

log √ x+hx

h





ñ ' ( x )=log(límh→0(1+ hx )

12h)

ñ ' ( x )=log(límh→0(1+ hx )

1h12

xx )=¿ log( límh→0(1+ h

x )xh)límh →0 ( 12x )¿

ñ ' ( x )=[ límh→0( 12x )] [ log (e) ]

ñ ' ( x )= log (e)2 x

i) q ( x )=ln √3 x (josue)solucion

q ' ( x )=límh→0

ln √3 (x+h )−ln√3 xh

=límh→0

ln√ 3 (x+h )3 x

h

q ' ( x )=límh→0ln(1+ h

x )12h=ln( límh→0(1+ h

x )12h )=ln(límh→0(1+ h

x )1h12

xx )

q ' ( x )=ln(límh→0(1+ hx )

xh )límh →0( 12x )=[ límh→0( 12 x )] [ ln(e)]





q ' ( x )= 12x

j) r ( x )=sech2(x ) (Porfi)solucion

sech(x )= 2

ex+e−x

r ' ( x )=límh→0( 4

(e( x+h)+e−( x+h ))2− 4

(e x+e− x )2 ) 1h

r ' ( x )=límh→0( 4 (e x+e−x )2−4 (e ( x+h)+e (−x+h ) )2

(e (x +h)+e−( x+h ) )2∗(e x+e−x )2 ) 1hr ' ( x )=4 lím

h→0( e2x+e−2x+2−e2 ( x+h )−e−2 ( x+h )−2

h (e( x+h)+e−( x+h ))2∗(ex+e− x)2 )

r ' ( x )=4 límh→0( e2x+ 1

e2 x−e2 ( x+h )− 1e2 ( x+h )

h (e ( x+h )+e−(x +h) )2∗(ex+e−x )2 )ANALISIS MATEMÁTICO III




r ' ( x )=4 límh→0(

e4 x+1e2x

− e4 ( x+h )+1e2( x+h)

h (e ( x+h)+e−( x+h ) )2∗(ex+e−x )2 )r ' ( x )=4 lím

h→0( e4 x e2 ( x+h )+e2 ( x+h )−e2x e4 ( x+h)−e2 x

h (e ( x+h)+e−( x+h ) )2∗(ex+e−x )2 e2 xe2 (x +h) )r ' ( x )=4 lím

h→0( e6x+4h+e2x+2h−e6 x+4h−e2 x

h (e ( x+h)+e−( x+h ) )2 (ex+e−x )2 (e2x )e2 ( x+h ) )r ' ( x )=4 lím

h→ 0( e (6 x+2h)(1−e2h)+e2x (e2h−1)

h (e ( x+h)+e−( x+h ) )2 (ex+e−x )2 (e2x )e2 ( x+h ) )=8 límh→ 0( (e2h−1)2h

(e2x−e6 x+4h ))

r ' ( x )=8 límh→0

( (e2h−1 )2h ) lím

h→0( e2x−e6x+4h

(e( x+h)+e−( x+h ))2 (ex+e−x )2 (e2 x)e2 ( x+h ))

r ' ( x )= 8(e2x−e6 x)

(ex+e−x )4 . e4 x=16(e−2 x−e2x )

2 (ex+e−x )4 .





r ' ( x )=( e

−2x−e2x

2)

( ex+e− x

2 )4

.

=−senh(2 x)cosh4 x

=−2 senh ( x ) .cosh (x)

cosh4 x

r ' ( x )=−2 senh ( x ) ¿ ¿coshx

.1

cosh2 x=−2 sech2 ( x ) . tanh (x)

k) s ( x )=coth3 ( x ) (Porfi)solucion

como :coth3 ( x )= cosh3 (x)

senh3(x )

s ' ( x )=limh→0 [ cosh

3 (x+h)senh3(x+h)

−cosh3 (x )senh3(x) ] 1h

s ' ( x )=limh→0 [ cosh3( x+h)senh3 x−cosh3 x senh3(x+h)

h (senh3(x+h) senh3 x) ]





s ' ( x )=limh→0 [ (cosh ( x ) cosh (h )+se nh ( x ) senh (h ) )3 senh3 x−cosh3 x (senh (x ) cosh (h )+cosh ( x ) senh (h ) )3

h (senh3( x+h)senh3 x ) ]s ' ( x )=lim

h→0(cosh3 (x ) cosh3 (h )+3cosh2 ( x ) cosh2 (h ) . senh ( x ) senh (h )+3cosh ( x )cosh (h ) . senh2 ( x ) senh2 (h )+senh3 ( x ) senh3 (h ) ) senh3 x−cosh3 x (senh3 ( x )cosh3 (h )+3 senh2 (x ) cosh2 ( h ) .cosh ( x ) senh (h )+3 senh ( x )cosh (h ) .cosh2 senh2 (h )+cosh3 ( x ) senh3 (h ) )

Hacemos C=cosh () y S=senh ()

s ' ( x )=limh→ 0

[S3 x C3 ( x )C3 (h )+3S4 xC2 ( x ) C2 (h ) . S (h )+3C ( x )C (h ) . S5 ( x ) S2 (h )+S6 ( x ) S3 (h ) ]−[C3 xS3 ( x ) C3 (h )+3S2 (x ) C2 (h ) .C4 xS (h )+3S ( x ) C (h ) .C5( x)S2 (h )+C6 ( x ) S3 (h ) ]

s ' ( x )=limh→0 (−3S

2 ( x ) (S ( x )C2 (h ) S (h ) (−S2 ( x )+C2 ( x ) ))h (senh3(x+h)senh3 x )

+3CxCh.Sx S2h(S4 x−C4 x)

h (senh3(x+h) senh3 x)+

S3h(S5 ( x )−C5(x ))h (senh3(x+h)senh3 x) )

s' (x)=limh→0

−3S2 (x ) S ( x )C2 (h ) S (h )h (senh3(x+h)senh3 x )

+limh→0

3CxCh. Sx S2h(S4 x−C4 x )h (senh3(x+h)senh3 x )

+ lim h→0

S3h(S5 (x )−C5(x ))h (senh3(x+h) senh3 x)

s' (x )=−3( limh→0

Shh )∗( limh→0

S2 ( x )C2 ( h )C2 (x )(senh3 (x+h ) senh3 x ) )+3( limh→0

Shh )∗(limh→0

CxCh .Sx S2h (S4 x−C4 x )(senh3 ( x+h ) senh3 x ) )+3( limh→0

Shh )∗lim

h→0

S3h(S5 ( x )−C5( x))

h (senh3(x+h)senh3 x )






s' (x )=−3 S2 ( x )C2 ( x )12

(senh3 x senh3 x )+(1 ) (0 )+(1 ) (0 )=−3 senh2 (x ) cosh2 ( x )

(senh3 x senh3 x )=

−3cosh2 ( x )( senh2 x )

∗1

( senh2 x )

s' (x )=−3coth2(x)∗cosh2(x)

l) u ( x )=arccos ( x )(Porfi)solucion

y=arcos ( x )x=cos y

x= e iy+e−iy

22x= e2 iy+1

eiy

2 xe iy=e2iy+10=e2iy−2 xe iy+1Por la ecuación general

e iy=2 x±√4 x2−42

eiy=2x ±2√x2−12

→eiy=x+√ x2−1

iy ln e=ln ( x+√x2−1 )





y=1iln ( x+√x2−1 )

Demuestremos “y”

y '=limh→0 ( 1i )∗¿ lim

h→ 0( ln (x+h )+√ (x+h )2−1−ln x+√ x2−1

h )¿

y '=1i

limh→0

1

h (ln [ ( x+h )+√ ( x+h )2−1x+√ x2−1 ])

y '=1i

limh→0

1

h (ln [1+ ( x+h )+√ ( x+h )2−1−x−√x2−1

x+√ x2−1 ])y '=1

i

limh→0

1

h (ln [1+ (h+√ (x+h )2−1−√ x2−1) (h+√ (x+h )2−1+√ x2−1)x+√x2−1 (h+√ (x+h )2−1+√ x2−1) ])

m) C ( x )=arctanh ( x ) (heber)solucion





arctanh ( x )=12ln( 1+x1−x )

C ' ( x )=limh→0 ( 12 )∗¿ lim

h→0 ( ln( 1+x+h1−x−h )−ln ( 1+x

1−x )) 1h ¿

C ' ( x )=12limh→0

1h ( ln [ 1+ x+h

1−x−h1+x1−x

])=12 lim h→0

1h (ln (1+ x+h ) (1−x )

(1−x−h ) (1+x ) )

C ' ( x )=12lim

h→0

1h (ln (1+x+h−x−x2−xh )

(1−x−h+x−x2−xh ) )=12 lim h→0

1h (ln( 1+h−x2−xh

1−h−x2−xh ))C ' ( x )=1

2lim

h→0

1h (ln(1+ 1+h−x2−xh

1−h−x2−xh−1))

C ' ( x )=12lim

h→0ln(1+ 1+h−x2−xh−1+h+x2+xh

1−h−x2−xh )1h

C ' ( x )=12ln [ lim h→0(1+ 2h

1−h−x2−xh )1h.22.1−h− x2− xh1−h− x2− xh ]





C ' ( x )=12ln [ lim h→0(1+ 2h

1−h−x2−xh )(1−h− x2−xh

2h )]lim h →0( 21−h− x2− xh)

C ' ( x )=12lim

h→0( 2

1−h−x2−xh ) ln e

C ' ( x )=12lim

h→0( 2

1−h−x2−xh )Reemplazando cuando h=0

C ' ( x )=12 ( 2

1−(0)−x2−x (0))C ' ( x )=1

2 ( 2

1−x2 )C ' ( x )= 1

1−x2





n) D ( x )=arccoth ( x ) (carlos)solucion

arctanh ( x )=12ln( 1+x1−x )

D ' ( x )=limh→0 ( 12 )∗¿ lim

h→0 ( ln( 1+x+h1−x−h )−ln( 1+x

1−x )) 1h ¿

D ' ( x )=12limh→0

1h ( ln [ 1+x+h

1−x−h1+x1−x

])=12 lim h→0

1h ( ln (1+x+h ) (1−x )

(1−x−h ) (1+x ) )

D ' ( x )=12lim

h→0

1h ( ln (1+x+h−x−x2−xh )

(1−x−h+x−x2−xh ) )=12 lim h→0

1h ( ln( 1+h− x2− xh

1−h−x2−xh ))D ' ( x )=1

2lim

h→0

1h ( ln(1+ 1+h−x2−xh

1−h−x2−xh−1))





D ' ( x )=12lim

h→0ln(1+ 1+h−x2−xh−1+h+x2+xh

1−h−x2−xh )1h

D ' ( x )=12ln [ lim h→0(1+ 2h

1−h−x2−xh )1h.22.1−h−x2−xh1−h−x2−xh ]

D ' ( x )=12ln [ lim h→0(1+ 2h

1−h−x2−xh )(1−h−x2−xh

2h )]lim h→0 ( 21−h−x2−xh )

D ' ( x )=12lim

h→0( 2

1−h−x2−xh ) ln e

D ' ( x )=12lim

h→0( 2

1−h−x2−xh )Reemplazando cuando h=0

D ' ( x )=12 ( 2

1−(0)−x2−x (0))D ' ( x )=1

2 ( 2

1−x2 )D ' ( x )= 1

1−x2





II. DIFERENCIAlES1. Haciendo uso de la diferencial hallar lo siguiente:

a¿ 3√25 (yendi)

solucion

f ( x )= 3√xx0=27

f ' (x0 )h=f (x0+h )−f (x0 )f ' (x0 )h=f (27−2 )−f (27)

f (25 )=f (27 )+ f ' (27 )(−2)3√25=3+ 1

33√272

(−2)

3√25=3+−233





2.9240≅ 2.9259

b) 3√67 (yendi)

solucionf (x+∆ x)=f (64+3)f ( x )= 3√x

f ´ (x)=1

33√x2

f (64 )= 3√64 = 4

f ´ (64)=1

33√642 =

148

∴ f (x+∆ x )=f (x)+ f ´ (x)∆ xf (64+3)=f (64)+f ´ (64)(3)

f (64+3)=3√64+ 1

33√642

(3)

f (64+3 )=4+ 348

3√67≅ 6516

3√25≅ 4.0625 3√25=4.0615





c) 3√83 (yendi)

solucion

f ( x+∆x )=f (64+19)

f ( x )= 3√x= 3√64

f ' ( x )= 1

33√x2

= 13√(64)2

= 112

h=19

f ( x+∆x )=f ( x )+ f ' ( x )∗h

f (64+19 )=f (64 )+ f ' (64 )∗19

3√83= 3√64+ 13√642

∗19

3√83≅ 4+ 1948





3√83≅ 21148

≅ 4.395

d) 4√17 (josue)

solucion

f ( x+∆x )=f (16+1)

f ( x )= 4√x=4√16

f ' ( x )= 1

44√x3

= 14√(16)3

= 132

h=1

f ( x+∆x )=f ( x )+ f ' ( x )∗h

f (16+1 )=f (16 )+ f ' (16 )∗1

4√17= 4√16+ 14√163

∗1

4√17≅ 2+ 132





3√17≅ 6532

≅ 2.031

e) 4√15 (josue)

solucion

f ( x+∆x )=f (16−1)

f ( x )= 4√x=4√16

f ' ( x )= 1

44√x3

= 14√(16)3

= 132

h=−1

f ( x+∆x )=f ( x )+ f ' ( x )∗h

f (16+1 )=f (16 )+ f ' (16 )∗(−1)

4√15=4√16+ 14√163

∗(−1)





3√15≅ 2− 132

3√15≅ 6332

≅ 1.9687

f) sen(28) (josue)

solucion

Dónde: sen(30 °−2 ° ) ; x = 30° , h = 2°

f(x) = sen x

f”(X) = cos(x)

f(x +¿ h) = hf”(x) +¿ f(x)

f(30 °−2 °) =¿ 2°)f”(30°) +¿ f(30°)

Convirtiendo: 2°( π180°

¿ = π90

f(28°) = −π90

. √32

+¿ 12

f(28°) = −0.030+0.5

f (28 °)=0.469

g) cos (27) (porfi)





solucion

f ( x )=cos ( x)

x0=30

h=−3

f '( x0)h=f (x0+h )−f (x0 )

f (x0+h )=f (x0 )+ f ' (x0 )h

f (27 )=f (30 )+ f ' (30 )(−3)

f (27 )=cos (30 )+3 sen (30)

cos (27 )=√32

+ 3π180 (12 )

cos (27 )=0.89

h) tan(32) (porfi)





solucion

f ( x )=tan (x)

x0=30

f ' (x0 )h=f (x0+h )−f (x0)

f ' (x0 )h=f (30+2 )−f (30)

f (32 )=f (30 )+ f ' (30 )(2Π180

)

tan (32 )=tan (30 )+ tan (30)' ( 2Π180

)

tan (32)= 1

√3+sec2(30)( 2Π

180)

tan (32 )≅ 1

√3+ 2Π135

i ¿cot29 ° (porfi)

solucion





sea: x0=30 y h=1

f (x0+h )=f (x0 )+h f ' (x0)

f (29 )=f (30 )+h f ' (30)

f 29 °=cot (x0 )+hcot' (x0)

cot 29 °=cot (x0 )+hcsc2(x0)

cot 29 °=cot (30 )+csc2(30)

cot 29 °=√3+4( π180

)

cot 29 °≅ 1.0802

j ¿ sec62 (heber)

solucion

f (x+∆ x)=f (60+2)

f ( x )=secx

f ´ (x)=sec x . tan x

f (64 )=sec 60 = 2

f ´ (64)=sec60 . tan60 =2√3





∴ f (x+∆ x )=f (x)+ f ´ (x)∆ x

f (60+2)=f (60)+f ´ (60)(2)

f (60+2)=sec 60+sec 60 . tan 60(2)

f (60+2 )=2+2√3 π180

(2)

sec62≅ 2+4√3 π180

sec62 ° ≅ 2.1209

k ¿ ln o .ooooooo5

i ¿eo .9 (heber)

solucion

f (x+∆ x)=f (eo.8+0.23406218261)

f ( x )=ex

f ´ (x)=1

ex

∴ f (x+∆ x )=f (x)+ f ´ (x)∆ x

f (eo .8+0.23406218261)=f (eo .8)+f ´ (eo .8)(0.23406218261)





f (eo .8+0.23406218261)=eo .8+ 1

eo .8(0.23406218261)

f (eo.8+0.234062182612 )=2.22554092849+ 12.22554092849

(0.23406218261)

eo .9≅ 2.22554092849+0.1051709196

eo .9≅ 2.377250124

eo .9≅ 2.45960311116∎

m ¿arctan0.97 (carlos)

solucionf (x+∆ x)=f (1+(−0.03))f ( x )=arctan x

f ´ (x)=1

1+ x2

f (1 )=arctan (1) = 45

f ´ (1)=1

1+(1)2 =1

∴ f (x+∆ x )=f (x)+ f ´ (x)∆ x





f (1+(−0,03))=f (1)+f ´ (−0.03)

f (1+(−0,03))=45+ 1

1+ (45 )2.

π108

(−0.03)

f (1+(−0,03))=45− 0.03 π

180(1+ (45 )2)

hallar la diferencial de las siguientes funciones:

a¿ f (x , y )=3√ x y2+cos3 (xy )+exsen(xy) (yendi)solucion

∂ f∂ x

= ∂∂ x

( 3√ x y2 )+ ∂∂ x

(cos3 ( xy ) )+ ∂∂x

(e¿¿ xsen(xy ))¿

∂ f∂ x

= ∂∂ x

(x1/3 y2 /3 )+3cos2(xy ) ∂∂ xcos( xy) ∂

∂xxy+ ∂

∂ xxsen(xy )(e¿¿ xsen (x y ))¿

∂ f∂ x

=133√( y

x )2

−3cos2(xy )sen (xy) y+(sen ( xy )+xcos (xy ) y )(e¿¿ xsen( xy))¿

∂ f∂ x

=133√( y

x )2

−3 ycos2(xy )sen(xy )+(sen ( xy )+xcos (xy ) y )(e¿¿ xsen ( xy ))¿

∂ f∂ y

= ∂∂ y

( 3√x y2 )+ ∂∂ y

(cos3 (xy ))+ ∂∂ y

(e¿¿ xsen(xy ))¿





∂ f∂ y

= ∂∂ y

(x1 /3 y2/3 )+3cos2(xy ) ∂∂ ycos(xy ) ∂

∂ yxy+ ∂

∂ yxsen (xy)(e¿¿ xsen (xy ))¿

∂ f∂ x

=233√ xy−3cos2(xy )sen (xy) x+(xcos (xy )x )(e¿¿xsen (xy ))¿

∂ f∂ x

=233√ xy−3x cos2(xy)sen (xy )+(xcos (xy )x )(e¿¿xsen (xy ))¿

f ( x , y )=¿¿Suderivadaimplícita :

∂ f ( xy )=

−∂ f ( xy )∂ x

∂ f ( xy )∂ x

f ( x , y )=3√ x y2+c os3 ( xy )+e xsen(xy )

∂ f ( xy )=−13

3√( yx )

2

+3cos2 ( xy ) sen ( xy ) y− (sen (xy )+xcos(xy ) y)(e¿¿ xsen(xy))

233√ xy−3 xcos2(xy )sen(xy )+(xcos(xy )x )(e¿¿ xsen(xy ))¿

¿

b¿ g x , y=x y−2+ 3√cot(x2 y)4−x+ex3 tan (xy) (josue)





solucion

∂∂ x

dg(x , y)= ∂∂ x

[x

y2+ 3√cot(x2 y)4−x+ex3 tan (xy)¿dx +

∂∂ y

[ x

y2+ 3√cot(x2 y)4−x+ex3 tan (xy)¿dy

1.- Derivando ∂∂ x

⟹ ∂∂x

dg(x , y )= ∂∂ x

[ x

y2+ 3√cot(x2 y)4−x+ex3 tan (xy)¿dx

⟹ ∂∂x

dg(x , y )=¿[ x

y2¿+ ∂

∂ x¿]

+∂∂ x

¿]¿dx

⟹ ∂∂x

dg(x , y )= ¿]dx

⟹ ∂∂x

dg(x , y )= ¿]dx

⟹ ∂∂x

dg(x , y )= ¿]dx

⟹ ∂∂x

dg(x , y )= ¿]dx

⟹ ∂∂x

dg(x , y )= ¿]dx





⟹ ∂∂x

dg(x , y )= ¿]dx

⟹ ∂∂x

dg(x , y )= ¿]dx ∎

1I.- Derivando ∂∂ y

⟹ ∂∂ y

dg(x , y)= ∂∂ y

[ x

y2+ 3√cot(x2 y)4−x+ex3 tan (xy)¿dy

⟹ ∂∂ y

dg(x , y)=¿[ x

y2¿+ ∂

∂ y¿]

+∂∂ y

¿]¿dy

⟹ ∂∂ y

dg(x , y)= ¿]dy

⟹ ∂∂ y

dg(x , y)= ¿]dy

⟹ ∂∂ y

dg(x , y)= ¿]dy

⟹ ∂∂ y

dg(x , y)= ¿]dy

⟹ ∂∂ y

dg(x , y)= ¿]dy





⟹ ∂∂ y

dg(x , y)= ¿]dy

⟹ ∂∂ y

dg(x , y)= ¿]dy ∎

∴ ∂∂ y

df (x , y )=¿]dy

e ¿ j (x , y )=senh2 ( xy )+arcsen ( 3√sech2 (x √ y3) ) (porfi)

Diferencial

solucion

j ( x , y )= ∂∂x

[s enh2 ( xy )+arcsen ( 3√sech2 (x √ y3 ))]+ ∂∂ y

[ senh2 ( xy )+arcsen ( 3√sech2 (x √ y3 )) ]

⟹ ∂∂x

[s enh2 ( xy )+arcsen ( 3√sech2 (x √ y3 ) )]=2 senh ( xy )cosh ( xy ) . y+ 13√sech2 ( x√ y3 )√ 3√sech4 ( x√ y3 )−1

1

33√sech4 (x √ y3)

.2 sech ( x√ y3 ) (−sech (x √ y3) ) . tanh (x √ y3) (√ y3)

∂∂ x

[ senh2 ( xy )+arcsen ( 3√sech2 (x √ y3 )) ]=2 ysenh ( xy )cosh ( xy )−2√ y3 sech2 (x √ y3 ) . tanh (x√ y3 )

3 sech2 ( x√ y3 )√ 3√sech4 ( x√ y3 )−1





⟹ ∂∂ y

[ senh2 ( xy )+arcsen ( 3√sech2 (x √ y3 )) ]=2 senh ( xy ) cosh ( xy ) . x+ 13√sech2 ( x√ y3 )√ 3√sech4 ( x√ y3 )−1

1

33√sech4 (x √ y3)

.2 sech ( x√ y3 ) (−sech (x √ y3) ) . tanh (x √ y3) ( 32 x √ y )

∂∂ y

[senh2 ( xy )+arcsen( 3√sech2 ( x√ y3 ))]=2xsenh ( xy )cosh ( xy )−x √ y sec h2 (x √ y3 ) . tanh (x √ y3 )

sech2 (x √ y3 )√ 3√sech4 (x √ y3 )−1

∴dj ( x , y )=[2 ysenh ( xy ) c osh ( xy )− 2√ y3 sec h2 (x √ y3 ) . tanh ( x√ y3 )3 sech2 (x √ y3) √ 3√sech4 (x √ y3)−1 ]dx+[2 xsenh ( xy ) cosh ( xy )− x √ y sec h2 ( x√ y3 ) . tanh (x √ y3 )

sech2 (x √ y3 )√ 3√sech4 (x √ y3 )−1 ]dy

Derivada implícitaj ( x , y )=s enh2 (xy )+arcsen( 3√sech2 (x √ y3 )) (heber)

Solución:

∂∂ x

[ senh2 ( xy )+arcsen ( 3√sech2 (x √ y3 )) ]=2 ysenh ( xy )cosh ( xy )−2√ y3 se c h2 (x √ y3 ) . tanh (x √ y3 )

3 sech2 ( x√ y3 )√ 3√sech4 ( x√ y3 )−1

∂∂ y

[senh2 ( xy )+arcsen( 3√sech2 ( x√ y3 ))]=2xsenh ( xy )cosh ( xy )−x √ y sec h2 (x √ y3 ) . tanh (x √ y3 )

sech2 (x √ y3 )√ 3√sech4 (x √ y3 )−1





j' ( x , y )=s enh2 (xy )+arcsen( 3√sech2 (x √ y3 ))=−[ ∂

∂ xj ( x , y )]

[ ∂∂ y

j (x , y )]

∴ j ' ( x , y )=

−[2 ysenh ( xy ) cosh ( xy )− 2√ y3 sec h2 ( x√ y3 ) . tanh (x √ y3 )3 sech2 (x √ y3 )√ 3√sech4 (x √ y3 )−1 ]

[2 xsenh ( xy ) cosh ( xy )− x√ y sec h2 (x √ y3 ) . tanh (x √ y3 )sech2 (x √ y3 )√ 3√ sech4 (x √ y3 )−1 ]

∴ j ' ( x , y )=−[6 ysenh ( xy )cosh ( xy ) sech2 ( x√ y3 )√ 3√sech4 ( x√ y3 )−1−2√ y3 sec h2 (x √ y3 ) . tanh ( x√ y3 )][6 xsenh ( xy ) cosh (xy ) sech2 (x √ y3 )√ 3√sech4 (x √ y3 )−1−x √ y sec h2 (x √ y3 ) . tanh (x √ y3 )]

III.- MÁXIMOS Y MINIMOS

En las funciones siguientes hallar:

a) Intervalos de concavidadb) Máximos y minimos relativosc) Máximos yminimos absolutosd) Puntos de intersección con eje Xe) Puntos de inflexión





1) f ( x )=x (x2−4)(x2−36) (yendi)solucion

f ( x )=x (x2−4)(x2−36)f ( x )=(x3−4 x)(x2−36)f ( x )=x5−36x3−4 x3+144 x

f ( x )=x5−40 x3+144 x

i . f (x ) ´=5x4−120 x2+144 (heber)

f ( x ) ´=0 ⟹5 x4−120 x2+144=0

⟹(x−√ 120+√1152010 )(x+√ 120+√11520

10 )(x+√ 120−√1152010 )(x−√ 120−√11520

10)=0

⟹ ( x−4.77 ) (x+4.77 ) ( x+1.13 )(x−1.13)=0f ( x ) ´ ´=20 x3−240 x

f (4.77 ) ´ ´=20 (4.77 )3−240 (4.77 )=1025.8 ¿ 0 ; f tiene unmínimo

f (−4.77 ) ´ ´=20 (−4.77 )3−240 (−4.77 ) ¿−1025.8 ¿ 0 ; f tiene unmáximo

f (1.13 )´ ´=20 (1.13 )3−240 (1.13 ) ¿−242.3 ¿ 0 ; f tiene unmáximo

f (−1.13 )´ ´=20 (−1.13 )3−240 (−1.13 ) ¿242.3 ¿ 0 ; f tiene unmínimo





ii . f ( x )=x5−40x3+144 x (carlos)solucion

f (4.77 )= (4.77 )5−40 (4.77 )3+144 (4.77 )=−1184.97 ; ⟹(4.77 ;−1184.97) es elmínimorelativo

f (−4.77 )=(−4.77 )5−40 (−4.77 )3+144 (−4.77 ) ¿1184.97 ; ⟹(−4.77 ;1184.97) es elmáximorelativo

f (1.13 )=(1.13 )5−40 (1.13 )3+144 (1.13 ) ¿106.84 ; ⟹(1.13 ;106.84) es elmáximorelativo

f (−1.13 )=(−1.13 )5−40 (−1.13 )3+144 (−1.13 )=−106.84 ; ⟹(−1.13 ;−106.84) es elmínimorelativo

iii . puntode inflexión solucion

f ( x ) ´ ´=0⟹20 x3−240 x=0

⟹ x (x+√12 ) (x−√12 )=0

f ( x )=x5−40 x3+144 x

f (0 )=(0 )5−40 (0 )3+144 (0 ) = 0

f (√12 )=(√12 )5−40 (√12 )3+144 (√12 ) = -665.11

f (−√12 )=(−√12 )5−40 (−√12 )3+144 (−√12 ) = 665.11

∴ por lotanto (0 ;0 )es el punto de inflexión

por lo tanto (√12;−665.11)es el punto de inflexión

por lo tanto (−√12;665.11 ) esel puntode inflexión





iv . concavidad

<−∞ ;−√12> →x=−4⟹ f (−4 )´ ´=20 (−4 )3−240 (−4 )=−320<0concava haciaabajo

<−√12; 0> →x=−1⟹ f (−1 )´ ´=20 (−1 )3−240 (−1 )=220>0concavahacia arriba

< 0;√12> →x=1⟹ f (1 )´ ´=20 (1 )3−240 (1 )=−220<0concavahacia abajo

<√12;+∞> →x=4⟹ f (4 )´ ´=20 (4 )3−240 (4 )=320>0concavahacia arriba

∴ f esconcava hacia arribaen←√12 ;0>U<√12 ;+∞>¿

f esconcava haciaabajo en←∞ ;−√12>U<0 ;√12>¿

GRAFICO DE LA FUNCIÓN





2) g ( x )=(x2−9)( x2−81) (josue)Solución

g ( x )=(x2−9)( x2−81)g ( x )=x4−81x2−9 x2+729g ( x )=x4−90x2+729

i . g ( x )´=4 x3−180 x





g ( x )´=0 ⟹4 x3−180 x=0 ⟹ x (4 x¿¿2−180)=0¿ ⟹ x (x+√45 )(x−√45)=0

g ( x )´ ´=12x2−180

g (0 ) ´ ´=12 (0 )2−180=−180 ¿ 0 ; g tieneunmáximo

g (−√45 ) ´ ´=12 (−√45 )2−180 ¿360 ¿ 0 ; g tieneunmínimo

g (√45 ) ´ ´=12 (√45 )2−180 ¿360 ¿ 0 ; g tieneunmínimo

ii . g ( x )=x 4−90 x2+729 (porfi)

g (0 )= (0 )4−90 (0 )2+729=729 ; ⟹(0;729) es elmáximorelativo

g (−√45 )=(−√45 )4−90 (−√45 )2+729 ¿−1296 ; ⟹(−√45 ;−1296) es elmínimorelativo

g (√45 )=(√45 )4−90 (√45 )2+729 ¿−1296 ; ⟹(√45;−1296) es elmínimorelativo

iii . puntode inflexión (porfi)

g ( x )´ ´=0⟹12x2−180=0

⟹ ( x+√15 ) (x−√15 )=0

g ( x )=x4−90x2+729

g ( x )=x4−90x2+729





g (−√15 )=(−√15 )4−90 (−√15 )2+729 = -396

g (√15 )=(√15 )4−90 (√15 )2+729 = -396

∴ por lotanto (−√15 ;−396 )esel punto de inflexión

por lo tanto (√15;−396 )esel punto de inflexión iv . concavid ad (porfi)

g ( x )´ ´=0⟹12x2−180

<−∞ ;−√15> →x=−4⟹ g (−4 ) ´ ´=12 (−4 )2−180=12>0concava haciaarriba

<−√15 ;√15> →x=0⟹g (0 )´ ´=12 (0 )2−180=−180<0concava haciaabajo

<√15 ;+∞> →x=4⟹ g (4 ) ´ ´=12 (4 )2−180=12>0concavahacia arriba

∴ges concavahacia arribaen←∞ ;−√15>U<√15;+∞>¿

ges concavahacia abajoen←√15 ;√15>¿






3) h ( x )=x (x2−25)(x2−144) (heber)Solución

h ( x )=(x3−25 x)(x2−144)h ( x )=x5−144 x3−25x3+3600 x





h ( x )=x5−169 x3+3600x

i .h ( x ) ´=5 x4−507 x2+3600h ( x )´=0 ⟹5 x4−507 x2+3600=0

⟹(x−√ 507+√18504910 )( x+√ 507+√185049

10 )(x−√ 507−√18504910 )(x+√ 507−√185049

10)=0

⟹ ( x+9.68 ) ( x−9.68 ) ( x+2.77 )(x−2.77)=0

h ( x )´ ´=20 x3−1014 x

h (9.68 )´ ´=20 (9.68 )4−1014 (9.68)=8325.26 ¿ 0 ; h tieneunmínimo

h (−9.68 )´ ´=20 (−9.68 )4−1014 (−9.68) ¿−8325.26 ¿ 0 ;h tienemáximo

h (2.77 )´ ´=20 (2.77 )4−1014(2.77) ¿−2383.70 ¿ 0 ;h tieneunmáximo

h (−2.77 )´ ´=20 (−2.77 )4−1014 (−2.77)=2383.70 ¿ 0 ;h tieneunmínimo

ii . h ( x )=x5−169 x3+3600 x

h (9.68 )=(9.68 )5−169 (9.68 )3+3600 (9.68 )=−33449.87 ; ⟹(9.68;−33449.87) es elmínimorelativo

h (−9.68 )=(−9.68 )5−169 (−9.68 )3+3600 (−9.68 )=33449.87 ; ⟹(−9.68 ;33449.87) es elmáximorelativo

h (2.77 )=(2.77 )5−169 (2.77 )3+3600 (2.77 ) ¿6543.16 ; ⟹(2.77 ;6543.16) es elmáximorelativo

h (−2.77 )=(−2.77 )5−169 (−2.77 )3+3600 (−2.77 ) ¿−6543.16 ; ⟹(−2.77 ;−6543.16) es elmínimorelativo

iii . puntode inflexión





h ( x )´ ´=0⟹20 x3−1014 x=0⟹ x (20 x¿¿2−1014)=0¿

⟹ x (x+√ 50710 )(x−√ 50710 )=0h ( x )=x5−169 x3+3600x

h (0 )=(0 )5−169 (0 )3+3600 (0 )=0

h(√ 50710 )=(√ 50710 )5

−169(√ 50710 )3

+3600(√ 50710 )=−17073.1

h(−√ 50710 )=(−√ 50710 )5

−169(−√ 50710 )3

+3600(−√ 50710 )=17073.1

∴ por lotanto (0 ;0 )es el punto de inflexión

por lo tanto(√ 50710 ;−17073.1)esel punto de inflexión

por lo tanto(−√ 50710 ;17073.1)esel puntode inflexión

iv . concavidad





h ( x )´ ´=20 x3−1014 x

<−∞ ;−√ 50710 > →x=−8⟹g (−8 )´ ´=20 (−8 )3−1014 (−8 )=−2128<0concava haciaabajo

<−√ 50710 ;0> →x=−1⟹ g (−1 ) ´ ´=20 (−1 )3−1014 (−1 )=994>0concava haciaarri ba

<0 ;√ 50710 > →x=1⟹ g (1 )´ ´=20 (1 )3−1014 (1 )=−994<0concavahacia abajo

<√ 50710 ;+∞> →x=8⟹g (8 )´ ´=20 (8 )3−1014 (8 )=2128>0concavahacia arriba

∴hesconcavahacia arribaen←√ 50710 ;0>U<√ 50710 ;+∞>¿

hes concavahacia abajo en←∞;−√ 50710 >U <0 ;√ 50710 >¿










IV.- METODO DE INTEGRACION INDEFINIDA

1. METODO DE DESCOMPOSICION

a¿∫ dx

2x2+5 x+2 (yendi)

solucion

∫ dx

2 x2+5 x+2=∫ dx

(2x+1)(x+2)=12∫

dx

(2 x+12

)(x+2)

¿ 12∫ dx

(x−12)(x+2)

=

12∗−2

3∫

((x−12 )−( x+2 ))dx( x−12 )( x+2 )

=−13∫

((x−12 )− (x+2 ))dx(x−12 ) (x+2 )

¿−13∫

(x−12 )dx(x−12 ) (x+2 )

+ 13∫ ( x+2 ) dx

(x−12 ) ( x+2 )=−13∫ dx

( x+2 )+ 13∫ dx

(x−12 )

∫ dx

2 x2+5 x+2=−13ln|x+2|+ 1

3ln|x−12|+k





donde k=c1+c2

b¿∫ dx(3+x ) (4+x ) (yendi)

solucion

Solución: Simplificando el integrando∫ dx

(x+3)(3 x+4)=15∫ (3 x+4 )−(3x+9)dx

(x+3)(3 x+4)

∫ dx(x+3)(3 x+4)

=15∫

3 x+4dx(x+3)(3 x+4)

−35∫

x+3dx(x+3)(3 x+4)

∫ dx(x+3)(3 x+4)

=15∫

dx(x+3)

−35∫

dx(3 x+4)

∫ dx(x+3)(3 x+4)

=15ln|x+3|−3

5ln|3x+4|+k





c ¿∫ ¿¿ (josue)

solucion

∫¿¿

∫¿¿

∴ 12 x−3x2−12x+5

= 12 x−3(x−(6+√31))(x−(6−√31))

m=6+√31 n=6−√31

12x−3(x−m)(x−n)

= Ax−m

+ Bx−n

12 x−3=Ax−An+Bx−Bm

12 x−3=( A+B ) x−(An+Bm)

De donde: A+B=12 ; An+Bm=3

A=12−B ; A (6−√31 )+B (6+√31 )=3

A=12−( 372−69√3162

) ; (12−B ) (6−√31 )+B (6+√31 )=3





A=372+69√3162

; 72−12√31−6 B+B√31+6 B+B√31=3

B=372−69√3162

∫¿¿

∫¿¿

∫¿¿

∫¿¿

∫¿¿

d ¿∫ x3dx2 x2−3 x+5

(josue)

solucion

12∫(x+ 3

2− x+152(2x2−3 x+5)

)dx





12∫ xdx+∫ 32 dx−1

2∫x+15

2 x2−3 x+5dx

12∫ xdx+∫ 32 dx−1

2(∫ x

2x2−3 x+5dx+∫ 15

2 x2−3x+5dx)

2 x2−3x+5=(x−3+i √314

)(x+ 3+i √314

)

12∫ xdx+∫ 32 dx−1

2¿

+∫ 15

2x2−3 x+5dx¿

12 ( x

2

2+ 3 x2

−12 (∫ 1

(x−3+i √314

)+( 3+i √31

4)∫ 1

(x−3+i √314

)(x+ 3+i √314

)+15∫ 1

(x−3+i√314

)(x+ 3+i√314

)dx))

12¿





∫(x−3−i √31

4)

(x−3−i√314

)(x+ 3+i√314

)dx−∫

(x−3+ i√314

)

(x−3−i √314

)(x−3+ i√314

)

∫ 1

(x−3−i√314

)dx−∫ 1

(x−3−i √314

)

ln|(x−3−i √314

)|−ln|(x−3−i √314

)|12¿

e ¿∫ x2d x

(x+3)(x−4)(x+5) (porfi)

solucion

∴ x2

(x+3)(x−4)(x+5)= A

x+3+ B

x−4+ C

x+5





x2=A ( x−4 ) ( x+5 )+B ( x+3 ) ( x+5 )+C(x+3)(x−4)

x2=A (x2+5x−4 x−20 )+B (x2+5 x+3x+15 )+C(x2−4 x+3 x−12)

x2=A (x2+ x−20 )+B (x2+8x+15 )+C(x2−x−12)

x2=A x2+Ax−20 A+B x2+8Bx+15 B+C x2−Cx−12C

x2=( A+B+C ) x2+( A+8 B−C ) x+(15B−20 A−12C)

De donde:

(a)……. A+B+C=1 ; 13 A+13 B+13C=13 ……. (d)

(b)……. A+8 B−C=0 ; A+8 B−C=0

(c)…….. −20 A+15 B−12C=0 ; −20 A+15 B−12C=0

Sumando (d)+ (b)+ (c) tenemos: −6 A+36 B=13

B=13+6 A36

……. (e)

Sumando (a)+ (b): 2 A+9 B=1…………… (f)

Reemplazando (e) en (f): 2 A+9( 13+6 A36 )=1





A=−914

Reemplazando A en (e): B=

13+6 (−914

)

36

B=1663

Reemplazando A, B en (a): −914

+ 1663

+C=1

C=2518

∫ x2d x

(x+3)(x−4)(x+5)=∫( A

x+3)dx+∫( B

x−4)d x+∫( C

x+5)d x

∫ x2d x

(x+3)(x−4)(x+5)=−914

∫(dx

x+3)+ 1663

∫(dx

x−4)+ 2518

∫(dx

x+5)

∫ x2d x

(x+3)(x−4)(x+5)=−914ln(x+3)+ 16

63ln(x−4 )+ 25

18ln(x+5)+K





g¿∫ dx

(sen2 ( x )−1)(cos2 (x )−1) (heber)

solucion

∫ d x

(sen2 ( x )−1)(cos2 ( x )−1)=∫ dx

(−cos2 ( x ))(−sen2 ( x ))

∫ d x

(sen2 ( x )−1)(cos2 ( x )−1)=∫ d x

sen2 ( x )cos2 (x )

∫ d x

(sen2 ( x )−1)(cos2 ( x )−1)=∫ 4 dx

(2 sen (x)cos (x))2

∫ d x

(sen2 ( x )−1)(cos2 ( x )−1)=4∫ d x

sen2(2 x)

∫ d x

(sen2 ( x )−1)(cos2 ( x )−1)=4∫ csc2(2x )dx

∫ d x

(sen2 ( x )−1)(cos2 ( x )−1)=4 (−ctg (2x )

2 )+K





∫ d x

(sen2 ( x )−1)(cos2 ( x )−1)=−2ctg (2 x )+K

h¿∫ (x−2)dxx3−6 x2+11 x

=∫ (x−2)dxx (x−3−√2 i)(x−3+√2 i) (heber)

solucion

∫ xdxx (x−3−√2 i)(x−3+√2i)

−2∫ dxx (x−3−√2 i)(x−3+√2i)

Integrandoen partes

Primera parte

∫ xdxx (x−3−√2 i)(x−3+√2i)

=∫ xdxx (x−3−√2i)(x−3+√2i)

=∫ [ (x−3−√2 i )−(x−3+√2i)]dx(x−3−√2i)(x−3+√2 i)

¿− 12√2i∫

[ ( x−3−√2i )−(x−3+√2 i)]dx(x−3−√2 i)(x−3+√2i)





¿− 12√2i∫

(x−3−√2 i )dx( x−3−√2i ) (x−3+√2 i)

+ 12√2 i∫

(x−3+√2 i )dx( x−3−√2i ) (x−3+√2 i )

¿− 12√2i∫

dx(x−3+√2 i )

+ 12√2 i∫

dx(x−3−√2 i )

¿− 12√2i

ln|(x−3+√2 i)|+c1+1

2√2iln|( x−3−√2 i )|+c2

Segunda parte

2∫ dxx (x−3−√2i)(x−3+√2 i)

= 2−2√2 i∫

[ ( x−3−√2 i)−(x−3+√2 i)]dxx (x−3−√2i)(x−3+√2 i)

¿ 1−√2 i∫

[ (x−3−√2 i )−(x−3+√2i)]dxx (x−3−√2 i)(x−3+√2i)

¿ 1−√2 i∫

( x−3−√2 i)dxx( x−3−√2 i)(x−3+√2 i)

− 1−√2i∫

(x−3+√2 i)dxx (x−3−√2 i)(x−3+√2i)

¿− 1

√2 i∫dx

x (x−3+√2 i)+ 1

√2 i∫dx

x (x−3−√2i)

¿− 1√2 i(3−√2 i)∫

x−(x−3+√2i)dxx (x−3+√2i)

+ 1√2 i(3+√2i)∫

x−(x−3−√2i)dxx (x−3−√2i)





¿− 1

√2 i (3−√2i )∫dx

( x−3+√2 i )+ 1

√2i (3−√2 i )∫dxx

+ 1

√2 i (3+√2i )∫dx

(x−3−√2 i )− 1

√2 i(3+√2i)∫dxx

¿− 1

√2 i (3−√2i )ln|x−3+√2i|+c3+

1

√2i (3−√2 i )ln|x|+c 4+

1

√2 i (3+√2 i )ln|x−3−√2i|+c5−

1

√2 i (3+√2 i )ln|x|+c6

Rta :∫ ( x−2 )dxx3−6 x2+11 x

= −12√2 i

ln|( x−3+√2 i )|+ 12√2 i

ln|(x−3−√2i )|+ 1

√2i (3−√2 i )ln|x−3+√2 i|− 1

√2i (3−√2 i )ln|x|− 1

√2 i (3+√2 i )ln|x−3−√2i|+ 1

√2i (3+√2i )ln|x|+k

donde k=c1+c2+c3+c4+c5+c6

2. METODO DE INTEGRACION POR PARTES

c ¿∫ x5 ex3dx (yendi)





solucion

∫ x5 ex3dx=∫ x3 x2 ex3dx

∫ x5 ex3dx=13∫ x3d (ex3)

u = x3 dv = d (ex3)

du = 3x2dx v = ex3

∫ x5 ex3dx=13

[x3 ex3−3∫ x2 ex3dx ]

∫ x5 ex3dx=13

x3ex3−∫ x2 ex3dx

∫ x5 ex3dx=13

x3ex3−13∫ d (e¿¿x3¿)¿¿

∫ x5 ex3dx=13

x3ex3−13ex3

∫ x5 ex3dx=13ex3(x3−1)

d ¿∫ √ ln (x )x2dx (carlos)





solucion

∫¿¿

u=¿

du=12¿ v= x3

3

¿

¿

u=¿

du=−12

¿ v= x3

3

¿¿

¿¿





u=¿

du=−32

¿ v= x3

3

¿¿

f) ∫ ta n3 ( ln (x2 ))dx (josue)

solucion

sea:

u=tan3 ( ln (x2 )) dv=dx

du=3 tan2 ( ln ( x2 )) sec2 ( ln (x2 )) 2xx2

dx v=x

∴ ∫ ta n3 ( ln (x2 ))dx=xta n3 ( ln (x2 ))−6∫ tan2 ( ln (x2 )) sec2 (ln (x2 ))dx (1)





∫ ta n2 ( ln (x2 ))sec 2 ( ln (x2)) dx

Sea:

u=xtan ( ln (x2 )) dv=tan (ln (x2 )) sec2 ( ln (x2) )

xdx

du=tan ( ln (x2 ))+x sec2 (ln (x2 )) 2xx2

v=12∫ tan ( ln (x2)) d (tan ( ln (x2 )))dx

v=14tan2 ( ln (x2 ))

∴∫ ta n2 ( ln (x2 ))sec 2 ( ln ( x2 )) dx= x4tan3 ( ln ( x2 ))−1

4∫¿¿

∴∫ ta n2 ( ln (x2 ))sec 2 ( ln ( x2 )) dx= x4tan3 ( ln ( x2 ))−1

4∫¿¿

∴ 32∫ ta n2 ( ln (x2 ))sec 2 ( ln ( x2 )) dx= x

4tan3 ( ln ( x2 ))−1

4∫¿¿

∴∫ ta n2 ( ln (x2 ))sec 2 ( ln ( x2 )) dx= x6ta n3 (ln (x2 ))−1

6∫¿¿

Remplazamos en 1

∫ ta n3 ( ln (x2 ))dx=xta n3 (l n (x2) )−6∫ ta n2 ( ln (x2 ))sec 2 ( ln ( x2 )) dx





∫ ta n3 ( ln (x2 ))dx=xta n3 ( ln (x2 ))−6¿

∫ ta n3 ( ln (x2 ))dx=xta n3 ( ln (x2 ))−xtan3 ( ln (x2) )+∫¿¿

0=0

g¿∫ ln( ln ( 2x3 ))dx (josue)

solucion

De donde: u=ln ( ln( 2x3

)) ; d v=dx

du=

1

ln(2 x3 )

( 12 x3

)( 23)d x

; v=x

du=

1

ln(2 x3 )

( 32 x

)( 23)dx

du=

d x

x ln ( 2x3

)





∫ ln (ln (2 x3

))d x=x ln(ln( 2 x3

))−∫ x dx

x ln ( 2x3

)

∫ ln (ln (2 x3

))d x=x ln(ln( 2 x3

))−∫ d x

ln( 2x3

)

m=2 x3

→dm=2dx

3

∫ ln(ln (2 x3

))d x=x ln(ln( 2 x3

))−∫ 3dm

2 ln(m)

∫ ln (ln (2 x3

))d x=x ln(ln( 2 x3

))−32∫( ln(m))−1dm

∴Es una función circular

i ¿∫ x3d x

(x2−2)3 (porfi)





solucion

∫ x3d x

(x2−2)3=∫ x2 x dx

(x2−2)3

De donde: d v=x dx

(x2−2)3 ; u=x2→du=2 xd x

v=∫ x d x

(x2−2)3 ; m=x2−2→dm=2 x dx

v=∫ dm

2 (m)3

v=12∫m−3dm

v=12(m

−2

−2)

v=−1

4 (x2−2)2





∫ x3d x

(x2−2)3=

−( x)2

4 (x2−2)2+ 24∫ x dx

(x2−2)2

∫ x3dx

(x2−2 )3=

− (x )2

4 ( x2−2 )2+ 12∫ dm

2 ( m)2

∫ x3d x

(x2−2)3=

−( x)2

4 (x2−2)2+ 14∫m−2dm

∫ x3d x

(x2−2)3=

−( x)2

4 (x2−2)2+ 14 (m

−1

−1 )+K

∫ x3d x

(x2−2)3=

−( x)2

4 (x2−2)2−14 ( 1

(x2−2 ) )+K

∫ x3d x

(x2−2)3=−14 ( x2

(x2−2 )2+ 1

( x2−2 ) )+K

∫ x3d x

(x2−2)3=−12 ( x2−1

(x2−2 )2 )+K





o¿∫ ln ( x )x3

dx (heber)

solucion

u=ln (x )dv=dx

x3

du=1xdx v=∫ x−3dx v=

− x−2

2

∫ ln ( x )x3

d x= −x−2

2ln (x)+∫ x−2

2 ( 1x )dx

∫ ln ( x )x3

d x ¿−x−2

2ln (x )+ 1

2∫ x−3dx

∫ ln ( x )x3

d x =−x−2

2ln (x)+ 1

2( x

−2

−2)





∫ ln ( x )x3

d x=−x−2

2ln ( x )− 1

4 x2+k

∫ ln ( x )x3

d x=−ln ( x )2x2

− 1

4 x2+k

∫ ln ( x )x3

d x=1

2 x2(−ln ( x )−1

2)+k

∫ ln ( x )x3

d x=1

2 x2(−2 ln ( x )−1

2)+k

∫ ln ( x )x3

d x=−(2 ln ( x )+14 x2

)+k

3. METODO DE INTEGRACION TRIGONOMETRICA

m ¿∫ cosh2 ( x ) cosh2 (2 x ) tanh2 ( x ) dx (porfi)

solucion

∫cosh2 ( x )cosh2 (2x ) senh2 ( x )

cosh2 ( x )dx

∫cosh2 (2 x ) senh2 (x ) dx





∫cosh2 (2 x )( cosh (2 x )−12 )dx

∫cosh2 (2 x )( cosh (2 x )−12 )dx

12∫ [cosh3 (2x )−cosh 2(2x )]dx

12∫ [cosh2 (2x ) cosh (2x )]dx−1

2∫ cosh2 (2 x ) dx

12∫ [ (senh2 (2x )+1 )cosh (2 x )]dx−1

2∫( 1−cosh (4 x )2 )dx

12∫ senh2 (2 x ) cosh (2x )dx+ 1

2∫cosh (2x )dx−14∫ (1−cosh (4 x ) )dx

12∫ senh2 (2 x ) cosh (2x )dx+ 1

2∫cosh (2x )dx−14∫ dx+ 1

4∫cosh (4 x )dx

14∫ senh2 (2 x ) d (senh (2 x ))+ 1

2∫cosh (2x )dx−14∫ dx+ 1

4∫cosh (4 x )dx

senh3(2x )12

+c1+senh (2 x)

4+c2−

x4+c3+

senh (4 x )16

+c4





senh3(2x )12

+senh(2 x)

4− x4+senh (4 x )16

+k donde :c=c1+c2+c3+c4

n¿∫coth3 (2x ) senh2 (2 x ) dx (porfi)

solucion

∫ cosh3 (2x )

senh3 (2x )senh2 (2x ) dx

∫ cosh3 (2x )

senh (2 x )dx

∫ cosh2 (2x ) cosh (2 x )senh (2 x )

dx

∫ (senh2 (2x )+1 )cosh (2x )senh (2x )

dx

∫ senh(2x )cosh (2 x ) dx+∫coth (2x )dx

12∫ senh (4 x)dx+∫ coth (2 x )dx





cosh (4 x)4

+c1+ln ⌈ senh (2 x)⌉

2+c2

cosh (4 x)4

+ln ⌈ senh(2 x)⌉

2+C Donde :C=c1

ñ¿∫ senh3 ( x )cosh2 ( x ) sen2 (2x ) cos2(2x )dx (porfi)

solucion

∫ senh2 ( x ) senh(x )cosh2 ( x ) sen2 (2 x )cos2(2x )dx

∫( cosh (2 x )−12 )senh (x )(1−cosh (2 x )

2 )( 1−cosh (4 x )2 )(1+cosh (4 x )

2 )dx116∫ (cosh (2 x ) senh ( x )−senh (x )) (1−cos (4 x )−cosh (2x )+cos (4 x)cosh (2 x)) (1+cos (4 x))dx

116

∫(12 senh (2+1 ) x+ 12senh (2−1 ) x )(1−cos (4 x )−cosh (2x )+cos (4 x)cosh (2 x)) (1+cos (4 x))dx

132∫ ¿¿

132∫ ¿¿





132∫ ¿¿

∫¿¿

∫¿¿

∫¿¿

4. METODO DE COMPLETAR CUADRADOS

a¿∫ dx

x2−5x+2 (yendi)

solucion

x2−5 x+2=( x−52 )2

−254

+2=(x−52 )

2

−(√ 174 )2





∫ dx

( x−52 )2

−(√ 174 )2

∫ dx

u2−a2= 12aln|u−a

u+a |+c

Reemplazandoen la formula

∫ dx

( x−52 )2

−(√ 174 )2= 1

2(√ 174 )ln|( x−52 )−√ 174

(x−52 )+√ 174 |∫ dx

( x−52 )2

−(√ 174 )2= 1

√17ln|( 2 x−5

2 )−√172

(2 x−52 )+ √17

2|

∫ dx

( x−52 )2

−(√ 174 )2= 1

√17ln| x−5+√17

2

x−5−√172

|+k





b¿∫ (x+2)dx3x2+7 x−12

(josue)

solucion

Completando cuadrados

3 x2+7 x−12

x2+ 73

x−4

(x+ 76 )

2

−( 76 )2

−4

(x+ 76 )

2

−4936

−4

(x+ 76 )

2

−19336

Reemplazando en la integral





∫ dx

u2−a2= 12aln|u−a

u+a |+c

∫ (x+2)dx

( x+ 76 )2

−(√ 19336 )2

∫ d (x+2)

( x+ 76 )2

−(√ 19336 )2

∫ (x+2)dx3 x2+7 x−12

= 1

2√ 19336ln|(x+ 7

6 )−√ 19336( x+ 76 )+√ 19336 |+k

Donde: k=C

g¿∫ (x¿¿2−x+1)dxx2+12x

¿ (porfi)

solucion





∫ (x¿¿2−x+1)dxx2+12 x

=∫ x2dxx2+12x

−∫ xdxx2+12 x

+∫ dxx2+12 x

¿

∫ (x¿¿2−x+1)dxx2+12 x

=∫ xdxx+12

−∫ dxx+12

+∫ dxx2+12x

¿ ….completando cuadrados

∫ (x¿¿2−x+1)dxx2+12 x

=∫ (x+12)dxx+12

−12∫ dxx+12

−∫ dxx+12

+∫ dxx2+12 x+36−36

¿

∫ (x¿¿2−x+1)dxx2+12 x

=∫ dx−13 ln|x+12|+c1+∫ dx¿¿ ¿¿

∫ (x¿¿2−x+1)dxx2+12 x

=x+c1−13 ln|x+12|+c2+∫ dx¿¿ ¿¿

∫ (x¿¿2−x+1)dxx2+12 x

=x+c1−13 ln|x+12|+c2+112ln| x

x+12|+c3¿

∫ (x¿¿2−x+1)dxx2+12 x

=x−13 ln|x+12|+ 112ln| x

x+12|+k ¿

h¿∫ (x3−x )dxx 4−6 x2+4

(heber)

Solución:





∫ (x3−x)dxx4−6x2+4

=14∫ (4 x3−4 x )dx

x 4−6 x2+4

∫ (x3−x)dxx4−6x2+4

=14∫ (4 x3−4 x−8 x+8 x )dx

x4−6 x2+4

∫ (x3−x)dxx4−6x2+4

=14∫ (4 x3−12 x+8 x )dx

x4−6 x2+4

∫ (x3−x)dxx4−6x2+4

=14∫ (4 x3−12 x )dx

x4−6 x2+4+ 14∫ 8 xdx

x4−6 x2+4

completando cuadrados:

∫ (x3−x)dxx4−6x2+4

=14∫ d (x4−6 x2+4)

x 4−6 x2+4+∫ 2xdx

x4−6 x2+4

∫ (x3−x)dxx4−6x2+4

= ln|x 4−6 x2+4|+c1+∫ 2xdxx4−6 x2+4+5−5

∫ (x3−x)dxx4−6x2+4

=ln|x 4−6 x2+4|+c1+∫ 2 xdx¿¿¿ ¿

∫ (x3−x)dxx4−6x2+4

= ln|x 4−6 x2+4|+c1+∫ d( x¿¿2−3)

¿¿¿ ¿¿





∫ (x3−x)dxx4−6x2+4

= ln|x 4−6 x2+4|+c1+12√5

ln¿

∫ (x3−x)dxx4−6x2+4

=ln|x 4−6 x2+4|+ 12√5

ln ¿

i ¿∫ x3−2 xdxx4−4 x2+8

(carlos)

solucion

∫ x (x2−2)dx¿¿¿ ¿

Haciendo: u ¿ x2−2

du = 2xdx

∫ x3−2 xdxx4−4 x2+8

=12∫ udu

u2+4

∫ x3−2 xdxx4−4 x2+8

=12( 12)∫ 2udu

u2+4

∫ x3−2 xdxx4−4 x2+8

=14ln|u2+4|





∫ x3−2 xdxx4−4 x2+8

=14ln|(x2−2)2+4|

∫ x3−2 xdxx4−4 x2+8

=14ln|x4−4 x2+8|

METODO DE INTEGRACION COMPLEJA

a¿∫ sen6 (2x ) dx (Carlos)

solucion

∫ sen6 (2 x )=∫ (sen2 (2x ) )2 . (sen2 (2 x ) )dx

∫ sen6 (2 x )=∫( 1−cos4 x2 )

2

. (sen2 (2 x ) )dx

∫ sen6 (2 x )=∫ 14 (1−2cos 4 x+co s24 x ) (sen2 (2x ) )dx

¿ 14∫ sen2 (2 x ) dx−2

4∫ cos4 x sen2 (2x ) dx+ 14∫ cos24 x sen2 (2 x )dx





parteA=14∫ sen2 (2 x ) dx

parteB=−24 ∫cos 4 x sen2 (2x ) dx

parteC=14∫ cos24 x sen2 (2 x )dx

RESOLVIENDO LA PARTE “A”

¿ 14∫( 1−cos 4 x

2 )dx=18∫ (1−cos 4 x ) dx=18∫1dx−1

8∫cos 4 x dx

14∫ sen2 (2 x ) dx=1

8 ( x−14 sen 4 x)+k

RESOLVIENDO LA PARTE “B”

−24∫cos 4 x sen2 (2 x ) dx=−2

4∫cos 4 x ( 1−cos 4 x

2 )dx−14 ∫ (cos 4 x−co s24 x )dx=−1

4(∫cos 4 x dx−∫co s24 xdx )

−14 ( 14 sen 4 x−∫ co s24 x dx )=−1

4 ( 14 sen4 x−∫( 1+cos8 x2 )dx )





−14 ( 14 sen 4 x−

12∫ 1dx+

12∫cos 8x dx )=−1

4 ( 14 sen4 x−12 (x+

18sen8 x))

−24∫cos 4 x sen2 (2 x ) dx=−1

4 ( 14 sen 4 x−12x− 116

sen8 x)+k

RESOLVIENDO LA PARTE “C”

14∫ co s24 x sen2 (2 x ) dx=1

4∫ co s24 x ( 1−cos 4 x

2 )dx18

(∫ co s24 x dx−∫co s34 xdx )=18 (∫( 1+cos 8x2 )dx−∫ cos4 x∗(1−sen24 x )dx)

¿ 18 (12 (∫1dx+∫ c os8 x dx)−∫cos 4 x dx+∫ s en24 x d (sen4 x ))

¿ 18 (12 (x+ 1

8sen8 x)−14 sen 4 x+ 1

4sen34 x3 )

14∫ co s24 x sen2 (2 x ) dx=1

8 ( 12 x+ 116

sen8 x−14

sen4 x+ 14

s en34 x3 )+k

Respuesta:

∫ sen6 (2 x )dx=18 (x− 14 sen 4 x)− 14 ( 14 sen 4 x−1

2x− 116

sen8 x )





+18 ( 12 x+ 1

16sen8 x−1

4sen 4 x+ 1

4se n34 x3 )+k

b¿∫cot5 (3 x ) dx (josue)

solucion

∫cot 2 (3 x ) cot3 (3 x ) dx

∫(cs c23 x−1)cot3 (3 x ) dx

∫ csc23 x cot3 (3x ) dx−∫cot3 (3 x ) dx

−13 ∫co t 33 x d (csc 3 x)dx−∫(csc23x−1)cot (3 x )dx





−112

c ot 43 x−∫ csc23 xcot (3x ) dx+∫cot (3 x )dx

∫cot5 (3 x ) dx=−112

c ot 43 x+ 16cot 23 x+ 1

3ln sen3 x+k

q¿∫ excosn ( x ) dx= ex

−2(2 i)n−1 {[(n0) (nsen (nx )+cos (nx ) )1+n2 ]−(n1)[ (n−1) (sen ((n−1)x )+cos ((n−1) x ) )

1+(n−1 )2 ]+(n2) [ (n−2 ) sen ( (n−2 ) x )+cos ( (n−2 ) x )1+(n−1 )2

−…−( nn2 )2n ]}

A) Analizando la integral cuando n=imparSabemosquesen (ax )= eaxi−e−axi

2i; a=1

reemplazandoa la integral.

¿∫ex ( exi−e−xi

2 i )n

dx

utilizandobinomio denewton





(a−b )n=(n0)an−(n1)an−1b+(n2)an−2b2−…+( nn−1)abn−1−(nn)bn

senn ( x )=

(n0) (enxi−e−nxi )−(n1) (e (n−1 ) xi−e−(n−1) xi )+…−( nn2 )2n−1

Sabemosque(n0)=(nn);(n1)=( nn−1)…

¿ 1(2 i)n−1

∫ex [(n0)( enxi−e−nxi

2 i )+(n1)( e( n−1) xi−e−(n−1 ) xi

2 i )+…+( nn2 )]dx¿ 1(2 i)n−1

∫ex [(n0)sen (nx )+(n1)sen ( (n−1 ) x )+(n2)sen ( ( n−2 ) x )+…+( nn2 )sen ( n2 )]dx

¿(n0)

(2 i)n−1∫ex sen (nx )dx+

(n1)2n−1∫ex sen ( (n−1 ) x )dx+¿

(n2)2n−1∫ex sen ( (n−2 ) x )dx+…+

( nn2 )2n−1∫ ex sen( n2 )dx¿





sabemosque∫eax sen (bx ) dx=eax (asen (bx )+bcos (bx ) )

a2+b2

r ¿∫ ex senn ( x ) dx (porfi)B) solucionC) Analizando la integral cuando n= par

Sabemosquesen (ax )= eaxi−e−axi

2i; a=1

reemplazandoa la integral.

¿∫ex ( exi−e−xi

2 i )n

dx

utilizandobinomio denewton





(a−b )n=(n0)an−(n1)an−1b+(n2)an−2b2−…−( nn−1)abn−1+(nn)bn

senn ( x )=

(n0) (enxi−e−nxi )−(n1) (e (n−1 ) xi+e−( n−1 ) xi )+…−( nn2 )¿¿

Sabemosque(n0)=(nn);(n1)=( nn−1)…

¿ 1−2(2 i)n−2

∫ex [(n0)( enxi+e−nxi

2 )−(n1)( e( n−1 ) xi+e−( n−1) xi

2 )−…( nn2 )]dx¿ 1−2(2 i)n−2

∫ex [(n0)cos (nx )−(n1)cos ( (n−1 ) x )+(n2)cos ( (n−2 ) x )−…−( nn2 )]dx

¿(n0)

−2(2 i)n−2∫ex cos (nx ) dx−

(n1)−2(2 i)n−2

∫ex cos ( (n−1 ) x)dx+¿(n2)

−2(2 i)n−2∫ex cos ( (n−2 ) x )dx−…−

( nn2 )−2(2i)n−2

∫ exdx ¿

sabemosque∫eaxcos (bx )dx=eax (bsen (bx )+acos (bx ) )

a2+b2






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