NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.4

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 1.
\(\frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 } \)
Solution:
Let x3 = t ⇒ 3x²dx = dt
\(\int { \frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 } dx } =\int { \frac { dt }{ { t }^{ 2 }+1 } } ={ tan }^{ -1 }t+C\)
= tan-1 (x3) + C

Question 2.
\(\frac { 1 }{ \sqrt { 1+{ 4x }^{ 2 } } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 1

Question 3.
\(\frac { 1 }{ \sqrt { { (2-x) }^{ 2 }+1 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 2

Question 4.
\(\frac { 1 }{ \sqrt { 9-{ 25x }^{ 2 } } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 3

Question 5.
\(\frac { 3x }{ 1+{ 2x }^{ 4 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 4

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 6.
\(\frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } } \)
Solution:
put x³ = t,so that 3x²dx = dt
\(\int { \frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } } dx } \quad =\frac { 1 }{ 3 } \int { \frac { dt }{ 1-{ t }^{ 2 } } \quad =\frac { 1 }{ 6 } log } \left| \frac { 1+t }{ 1-t } \right| +C\)
\(=\frac { 1 }{ 6 } log\left| \frac { 1+{ x }^{ 3 } }{ 1-{ x }^{ 3 } } \right| \) + C

Question 7.
\(\frac { x-1 }{ \sqrt { { x }^{ 2 }-1 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 5

Question 8.
\(\frac { { x }^{ 2 } }{ \sqrt { { x }^{ 6 }+{ a }^{ 6 } } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 6

Question 9.
\(\frac { { sec }^{ 2 }x }{ \sqrt { { tan }^{ 2 }x+4 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 7

Question 10.
\(\frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 8

Question 11.
\(\frac { 1 }{ { 9x }^{ 2 }+6x+5 } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 9

Question 12.
\(\frac { 1 }{ \sqrt { 7-6x-{ x }^{ 2 } } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 10

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 13.
\(\frac { 1 }{ \sqrt { (x-1)(x-2) } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 11

Question 14.
\(\frac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 12

Question 15.
\(\frac { 1 }{ \sqrt { (x-a)(x-b) } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 13

Question 16.
\(\frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 14

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 17.
\(\frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } } \)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 15

Question 18.
\(\frac { 5x-2 }{ 1+2x+{ 3x }^{ 2 } } \)
Solution:
Let 5x – 2 = A(6x + 2) + B … (1)
Equating the coefficients of x, we get
6A = 5 ⇒ A = \(\frac { 5 }{ 6 }\)
Equating the constant terms, we get
2A + B = – 2 ⇒ B = – 2 – 2A
= – 2 – \(\frac { 5 }{ 3 }\) = \(\frac { – 11 }{ 3 }\)
∴ (1) → 5x – 2 = \(\frac { 5 }{ 6 }\) (6x + 2) – \(\frac { 11 }{ 3 }\)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 16

Question 19.
\(\frac { 6x+7 }{ \sqrt { (x-5)(x-4) } } \)
Solution:
∫\(\frac { 6x+7 }{ \sqrt { (x-5)(x-4) } } \)dx = \(\int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x\)
Let 6x + 7 = A\(\frac { d }{ dx }\)(x² – 9x + 20) + B
6x + 7 = A(2x – 9) + B
Equating coefficients of x, we get
2A = 6 ⇒ A = 3
Equating the constant terms,
we get 7 = – 9A + B
= B = 7 + 9A ⇒ B = 7 + 27 = 34
i.e, 6x + 7 = 3(2x – 9) + 34
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 17

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 20.
\(\frac { x+2 }{ \sqrt { 4x-{ x }^{ 2 } } } \)
Solution:
Let x + 2 = A\(\frac { d }{ dx }\)(4x – x²) + B
x + 2 = A(4 – 2x) + B.
Equating coeffleints of x and constants an both sides, we get
– 2A = 1 ⇒ A = \(\frac {- 1 }{ 2 }\) and 4A + B = 2
⇒ B = 2 – 4A
⇒ B = 2 – 4\(\frac {- 1 }{ 2 }\) = 4
i.e., x + 2 = \(\frac {- 1 }{ 2 }\)(4 – 2x) + 4
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 18

Question 21.
\(\frac { x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } } \)
Solution:
∫\(\frac { x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } } \)dx
Let x + 2 = A\(\frac { d }{ dx }\)(x² + 2x + 3) + B
x + 2 = A(2x + 2) + B
Equating coefficients of x and constant terms on both sides, we get
2A = 1 ⇒ A = \(\frac { 1 }{ 2 }\) and 2A + B = 2
⇒ B = 2 – 4A
⇒ B = 2 – 2\(\frac { 1 }{ 2 }\) = 2 – 1 = 1
i.e., x + 2 = \(\frac { 1 }{ 2 }\)(2x + 2) + 1
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 19

Question 22.
\(\frac { x+3 }{ { x }^{ 2 }-2x-5 } \)
Solution:
Let x + 3 = A(2x – 1) + B
Equating coeffleints of x and constants an both sides, we get 2A = 1 ⇒ A = \(\frac { 1 }{ 2 }\)
Equating the constant terms, we get – 2A + B = 3 ⇒ B = 4 ∴ A = \(\frac { 1 }{ 2 }\), B = 4
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 20

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 23.
\(\frac { 5x+3 }{ \sqrt { { x }^{ 2 }+4x+10 } } \)
Solution:
Let 5x + 3 = A\(\frac { d }{ dx }\)(x² + 4x + 10) + B
5x + 3 = A(2x + 4) + B
Equating the coefficients ofx, we get
2A = 5 or A = \(\frac { 5 }{ 2 }\)
Equating the constant terms, we get
4A + B = 3 ⇒ B = 3 – 4A
= 3 – \(\frac { 4×5 }{ 2 }\) = – 7
∴ 5x + 3 = \(\frac { 5 }{ 2 }\)(2x + 4) – 7
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 21

Question 24.
\(\int { \frac { dx }{ { x }^{ 2 }+2x+2 } equals } \)
(a) xtan-1(x+1)+c
(b) (x+1)tan-1x+c
(c) tan-1(x+1)+c
(d) tan-1x+c
Solution:
(b) (x+1)tan-1x+c
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 22

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4

Question 25.
\(\int { \frac { dx }{ \sqrt { 9x-{ 4x }^{ 2 } } } equals } \)
(a) \(\frac { 1 }{ 9 } { sin }^{ -1 }\left( \frac { 9x-8 }{ 8 } \right) +c\)
(b) \(\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 8x-9 }{ 9 } \right) +c\)
(c) \(\frac { 1 }{ 3 } { sin }^{ -1 }\left( \frac { 9x-8 }{ 8 } \right) +c\)
(d) \({ sin }^{ -1 }\left( \frac { 9x-8 }{ 9 } \right) +c\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 23

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