1𝑑𝑥 = 𝑥 + 𝑐, 𝑐 ∈ ℝ

𝑥𝛼𝑑𝑥 =𝑥𝛼+1

𝛼+1+ 𝑐,

𝑐 ∈ ℝ , 𝛼 ∈ ℝ\{0,−1}

1

𝑥𝑑𝑥 = ln |𝑥| + 𝑐, 𝑐 ∈ ℝ

𝑒𝑥𝑑𝑥 = 𝑒𝑥 + 𝑐, 𝑐 ∈ ℝ

sin 𝑥 𝑑𝑥 = −cos 𝑥 + 𝑐, 𝑐 ∈ ℝ

cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐, 𝑐 ∈ ℝ

𝑓

𝐼𝐹

𝑓𝐼 𝐹

𝐼𝐹′ 𝑥 = 𝑓 𝑥 , ∀𝑥 ∈ 𝐼

𝐼

𝑓

𝑓 𝑥 = 1 , 𝑥 ≤ 00 , 𝑥 > 0

0 ∈ 𝐼 𝑓𝐼

𝑥𝛼𝑑𝑥𝛼

• 𝛼 = 0

1𝑑𝑥 = 𝑥 + 𝑐, 𝑐 ∈ ℝ

• 𝛼 = −1

1

𝑥𝑑𝑥 = ln|𝑥| + 𝑐, 𝑐 ∈ ℝ

•

𝑥𝛼𝑑𝑥 =𝑥𝛼+1

𝛼 + 1+ 𝑐, 𝑐 ∈ ℝ , 𝛼 ∈ ℝ\{0, −1}

1

𝑥𝑑𝑥

1

𝑥𝑑𝑥 =

𝑙𝑛𝑥 + 𝑐, 𝑐 ∈ ℝ

𝐼 ⊂ ℝ+ 1

𝑥𝑑𝑥 = 𝑙𝑛𝑥 + 𝑐, 𝑐 ∈ ℝ

𝐼 ⊂ ℝ− 1

𝑥𝑑𝑥 = ln −𝑥 + 𝑐, 𝑐 ∈ ℝ

𝑓 𝑓 𝑥 =1

𝑥

ℝ ∖ 0 𝑔 𝑔 𝑥 = 𝑙𝑛 𝑥0,+∞ .

𝑘𝑓 𝑥 𝑑𝑥 = 𝑘 𝑓(𝑥)𝑑𝑥

𝑓 𝑥 + 𝑔(𝑥) 𝑑𝑥 = 𝑓(𝑥)𝑑𝑥 + 𝑔(𝑥)𝑑𝑥

𝑢′ 𝑥 𝑓 𝑢 𝑥 𝑑𝑥 = 𝐹 𝑢 𝑥 + 𝑐, 𝑐 ∈ ℝ

𝐹 𝑓

a) 3𝑥𝑑𝑥 = 3 𝑥𝑑𝑥 = 3𝑥2

2+ 𝑐, 𝑐 ∈ ℝ

b) 3(2𝑥 − 1)2𝑑𝑥 = 3

22(2𝑥 − 1)2𝑑𝑥 =

3

2 2(2𝑥 − 1)2𝑑𝑥 =

=3

2×(2𝑥−1)3

3+ 𝑐, 𝑐 ∈ ℝ =

2𝑥−1 3

2+ 𝑐, 𝑐 ∈ ℝ

c) tg𝑥𝑑𝑥 = sin𝑥

cos𝑥𝑑𝑥 = −

−sin𝑥

cos𝑥𝑑𝑥 = − ln cos𝑥 + 𝑐, 𝑐 ∈ ℝ

d) 𝑥

2𝑑𝑥 =

𝑥

2𝑑𝑥 =

1

2 𝑥

1

2𝑑𝑥 =1

2×𝑥32

3

2

+ 𝑐, 𝑐 ∈ ℝ =2𝑥3

3+ 𝑐,

𝑐 ∈ ℝ

a) 𝑥3 + 𝑥2 + 1 𝑑𝑥

b) 𝑒3𝑥𝑑𝑥

c) 2𝑥𝑒𝑥2𝑑𝑥

d) 𝑥3 𝑥4 + 1 8𝑑𝑥

e) sin𝑥 cos2𝑥𝑑𝑥

f) ∗ cos3𝑥𝑑𝑥

g) ∗∗ ln (3𝑥)

𝑥𝑑𝑥

a)𝑥4

4+𝑥3

3+ 𝑥 + 𝑐, 𝑐 ∈ ℝ

b)𝑒3𝑥

3+ 𝑐, 𝑐 ∈ ℝ

c) 𝑒𝑥2+ 𝑐, 𝑐 ∈ ℝ

d)𝑥4+1

9

36+ 𝑐, 𝑐 ∈ ℝ

e) −cos3𝑥

3+ 𝑐, 𝑐 ∈ ℝ

f) sin𝑥 −sin3𝑥

3+ 𝑐, 𝑐 ∈ ℝ

g)ln (3𝑥) 2

2+ 𝑐, 𝑐 ∈ ℝ

𝑎, 𝑏 , (𝑎 ≤ 𝑏) 𝑓

𝑎 𝑏 𝑓 𝑥 𝑑𝑥𝑏

𝑎

𝑥 = 𝑎 𝑥 = 𝑏𝑓

𝑓 𝑥 𝑑𝑥3

1

• 𝑏 < 𝑎 𝑓 𝑥 𝑑𝑥𝑏

𝑎= − 𝑓 𝑥 𝑑𝑥

𝑎

𝑏

• 𝑓(𝑥) ≤ 0 𝑎, 𝑏 , (𝑎 ≤ 𝑏)

𝑓 𝑥 𝑑𝑥𝑏

𝑎

𝑥 = 𝑎𝑥 = 𝑏 𝑓

− 𝑓 𝑥 𝑑𝑥3

1

𝑓 𝑎, 𝑏 ,

(𝑎 < 𝑏) 𝐹𝑎 𝑎, 𝑏

𝐹𝑎 𝑥 = 𝑓 𝑦 𝑑𝑦𝑥

𝑎

𝑓 𝑎, 𝑏 𝐹𝑎 𝑎 = 0

𝑓 𝑎, 𝑏 ,

(𝑎 < 𝑏)

𝑓 𝑥 𝑑𝑥𝑏

𝑎

= 𝐹(𝑥) 𝑎𝑏 = 𝐹 𝑏 − 𝐹(𝑎)

𝐹 𝑓 𝑎, 𝑏

𝑓 𝑎, 𝑏 ,

(𝑎 < 𝑏) 𝐹 𝑓 𝑎, 𝑏

𝑓 𝑥 𝑑𝑥𝑏

𝑎= 𝐹𝑎 𝑏

𝐹 𝑓 𝐹 = 𝐹𝑎 + 𝑐

𝑐 ∈ ℝ

𝐹 𝑏 − 𝐹 𝑎 = 𝐹𝑎 𝑏 + 𝑐 − 𝐹𝑎 𝑎 + 𝑐 = 𝐹𝑎 𝑏 =

𝑓 𝑥 𝑑𝑥𝑏

𝑎

a) 3𝑥𝑑𝑥 =3𝑥2

2 1

3

=3×32

2−3×12

2

3

1=24

2= 12

b) (2𝑥 + 𝑒2𝑥+1)𝑑𝑥0

2=

= 2 𝑥𝑑𝑥0

2+1

2 2𝑒2𝑥+1𝑑𝑥0

2

= 2𝑥2

2 2

0

+1

2𝑒2𝑥+1 2

0

= 2(0 − 2) +1

2𝑒1 − 𝑒5

= −4 +𝑒−𝑒5

2

0 < 2

16

3

1

2

5

𝑒𝑥 + 𝑒−𝑥 𝑑𝑥ln3

−ln3

sin(2𝑥) 𝑑𝑥

𝜋3

𝜋6

∗ 𝑥 + 1 𝑑𝑥2

−2

𝑒2𝑦𝑑𝑦𝑥

0

𝑓 𝑥 = 𝑒2𝑥 𝑒2𝑦𝑑𝑦𝑥

0= 𝐹0(𝑥)

𝐹0′(𝑥)

𝐹0(𝑥) 𝑒2𝑦𝑑𝑦

𝑥

0

=𝑒2𝑦

20

𝑥

=𝑒2𝑥

2−1

2

𝐹0′(𝑥) 𝑒2𝑥

tan 𝑦3 𝑑𝑦0

2𝑥

𝑓 𝑥 = tan 𝑥3 tan 𝑦3 𝑑𝑦𝑥

0= 𝐹0(𝑥)

𝐹0′ 𝑥 = tan 𝑥3

𝐹0 𝑥 = tan 𝑦3 𝑑𝑦

𝑥

0

⇔ −𝐹0 𝑥 = tan 𝑦3 𝑑𝑦

0

𝑥

⇔

⇔ −𝐹0 2𝑥 = tan 𝑦3 𝑑𝑦

0

2𝑥

tan 𝑦3 𝑑𝑦0

2𝑥

′

= −𝐹0 2𝑥 ′ = −𝐹0′ 2𝑥 × 2 = −2tan 2𝑥

3

a) cos(𝑦2) 𝑑𝑦𝑥2

0

b) ∗∗ 1 − 𝑦2𝑑𝑦cos 𝑥

sin 𝑥

a) 2𝑥 cos(𝑥4)

b) −sin 𝑥 | sin 𝑥 | − cos 𝑥 | cos 𝑥 |

[𝐴𝐵𝐶𝐷]

𝐴 0, 2 , 𝐵 3, 1 , 𝐶 5, 5

𝐷 0, 4

𝐷𝐶 𝑦 =𝑥

5+ 4

𝐴𝐵 𝑦 = −𝑥

3+ 2

𝑥

5+ 4 𝑑𝑥

3

0

− −𝑥

3+ 2 𝑑𝑥

3

0

=42

5

𝐵𝐶

𝑦 = 2𝑥 − 5

18

5

12

𝑓 𝑥 = − 𝑥 − 2 + 4

𝑔 𝑥 = − 𝑥 − 1 + 2

Filipe Carvalho