# 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

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:

Question 3.
$$\frac { 1 }{ \sqrt { { (2-x) }^{ 2 }+1 } }$$
Solution:

Question 4.
$$\frac { 1 }{ \sqrt { 9-{ 25x }^{ 2 } } }$$
Solution:

Question 5.
$$\frac { 3x }{ 1+{ 2x }^{ 4 } }$$
Solution:

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:

Question 8.
$$\frac { { x }^{ 2 } }{ \sqrt { { x }^{ 6 }+{ a }^{ 6 } } }$$
Solution:

Question 9.
$$\frac { { sec }^{ 2 }x }{ \sqrt { { tan }^{ 2 }x+4 } }$$
Solution:

Question 10.
$$\frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } }$$
Solution:

Question 11.
$$\frac { 1 }{ { 9x }^{ 2 }+6x+5 }$$
Solution:

Question 12.
$$\frac { 1 }{ \sqrt { 7-6x-{ x }^{ 2 } } }$$
Solution:

Question 13.
$$\frac { 1 }{ \sqrt { (x-1)(x-2) } }$$
Solution:

Question 14.
$$\frac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } } }$$
Solution:

Question 15.
$$\frac { 1 }{ \sqrt { (x-a)(x-b) } }$$
Solution:

Question 16.
$$\frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } }$$
Solution:

Question 17.
$$\frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } }$$
Solution:

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) – $$\frac { 11 }{ 3 }$$

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

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

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

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

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

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

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:

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