Question 1: Obtain an integral (or anti - derivative) of the � u 2 1 u+ Answer 1: Suppose, u 3 = z 3u 2 du = dz J 3u 2 du-J u 6 +1 - z 2 +1 = tan- 1 z+ C =tan- 1 u 3 +C Question 2: Obtain an integral (or anti - derivative) of the 1 Jl+4u 2 Answer 2: Suppose, 2u = z 2 du= dz J 1 du - f J1+4u 2 - 2 = ½ [log j z+�IJ+c = ½ [log j 2u + ✓1 + 4u 2 IJ +C Question 3: Obtain an integral (or anti - derivative) of the 1 ✓(2) 2 +1 Answer 3: Suppose, 2-u = z -du= dz f 1 du =- f - 1 - dz ✓(2-) 2 +1 =- [log jz + ✓z 2 I] + C =- [ log 12-u + ✓(2-u) 2 + 1 IJ +C = log I ( 2)+� I + C Question 4: Obtain an integral (or anti - derivative) of the -- 1 - ✓9-25u 2 Answer 4: Suppose, 5u = z 5 du= dz J 1 du = f - 1 - dz ✓9-25u 2 5 9-z 2 = f-1 -dz 5 ✓3½2 = ½sin- 1 1 +c 1 _:_-1 S u , r, NCERT SOLUTIONS CLASS 12 MATHS CHAPTER-7 EXERCISE-7.3 myncertsolutions.com
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Question 1:
Obtain an integral (or anti - derivative) of the �u2
1 u+
Answer 1:
Suppose, u3 = z 3u2du = dz J 3u2 du-J_!k_u6+1 - z2+1= tan-1 z+ C =tan-1u3 +C
Question 2:
Obtain an integral (or anti - derivative) of the 1 Jl+4u2
Answer 2:
Suppose, 2u = z 2 du= dz J
1 du - _!_ f ____!!:!__J1+4u2 - 2 y'1+z2=½[logjz+�IJ+c = ½ [log j2u + ✓1 + 4u2 IJ + C
Question 3:
Obtain an integral (or anti - derivative) of the 1 ✓(2-11,)2+1
Answer 3:
Suppose, 2-u = z -du= dzf 1 du =- f -1- dz ✓(2-11,)2+1 .Jz2+i
=- [log jz + ✓z2 +TI] + C =-[log 12-u + ✓(2-u)2 + 1 IJ + C
= log I (2-11,)+� I + C
Question 4:
Obtain an integral (or anti - derivative) of the --1-✓9-25u2
Answer 4:
Suppose, 5u = z
5 du= dz
J 1 du = .! f -1- dz✓9-25u2 5 9-z2
= .! f-1-dz 5 ✓3½2
= ½sin-1 1 +c
1 _:_-1 Su , r,
NCERT SOLUTIONS CLASS 12 MATHS
CHAPTER-7 EXERCISE-7.3
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