# NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.7

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.7 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-7-ex-7-7/

## NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.7

Question 1.
$$\sqrt{4-x^{2}}$$
Solution:
$$\sqrt{4-x^{2}}$$ dx
= $$\int \sqrt{2^{2}-x^{2}} d x$$
= $$\frac{x}{2} \sqrt{2^{2}-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}+C$$
= $$\frac{x}{2} \sqrt{4-x^{2}}+2 \sin ^{-1} \frac{x}{2}+C$$

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

Question 3.
$$\sqrt { { x }^{ 2 }+4x+6 }$$
Solution:
$$\sqrt { { x }^{ 2 }+4x+6 }$$ dx
= $$\int \sqrt{x^{2}+4 x+4+6-4} d x=\int \sqrt{(x+2)^{2}+(\sqrt{2})^{2}} d x$$
= $$\frac{x+2}{2} \sqrt{(x+2)^{2}+(\sqrt{2})^{2}}+\frac{2}{2} \log \left|(x+2)+\sqrt{(x+2)^{2}+(\sqrt{2})^{2}}\right|+\mathrm{C}$$
= $$\frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\log \left|(x+2)+\sqrt{x^{2}+4 x+6}\right|+C$$

Question 4.
$$\sqrt{x^{2}+4 x+1}$$
Solution:
$$\sqrt{x^{2}+4 x+1}$$ dx
= $$\int \sqrt{x^{2}+4 x+4+1-4} d x=\int \sqrt{(x+2)^{2}-(\sqrt{3})^{2}} d x$$
= $$\frac{(x+2)}{2} \sqrt{(x+2)^{2}-(\sqrt{3})^{2}}-\frac{3}{2} \log \left|(x+2)+\sqrt{(x+2)^{2}-(\sqrt{3})^{2}}\right|+\mathrm{C}$$
= $$\frac{(x+2)}{2} \sqrt{x^{2}+4 x+1}-\frac{3}{2} \log \left|(x+2) \sqrt{x^{2}+4 x+1}\right|+C$$

Question 5.
$$\sqrt { 1-4x-{ x }^{ 2 } }$$
Solution:

Question 6.
$$\sqrt { { x }^{ 2 }+4x-5 }$$
Solution:
$$\sqrt { { x }^{ 2 }+4x-5 }$$dx
$$\int { \sqrt { { x }^{ 2 }+4x-5 } }$$ dx = $$\int { \sqrt { { (x+2) }^{ 2 }-{ (3) }^{ 2 } } }$$dx
= $$\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x-5 } -\frac { 9 }{ 2 } log|x+2+\sqrt { { x }^{ 2 }+4x-5 } |+c$$

Question 7.
$$\sqrt { 1+3x-{ x }^{ 2 } }$$
Solution:

Question 8.
$$\sqrt { { x }^{ 2 }+3x }$$
Solution:

Question 9.
$$\sqrt { 1+\frac { { x }^{ 2 } }{ 9 } }$$
Solution:

Question 10.
$$\int \sqrt{1+x^{2}}$$ dx is equal to
(a) $$\frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log|x+\sqrt { 1+{ x }^{ 2 } } |+C$$
(b) $$\frac { 2 }{ 3 } { \left( 1+{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+C$$
(c) $$\frac { 2 }{ 3 } x{ \left( 1+{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+C$$
(d) $$\frac { { x }^{ 2 } }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } { x }^{ 2 }log\left| x+\sqrt { 1+{ x }^{ 2 } } \right| +C$$
Solution:
(a) $$\frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log|x+\sqrt { 1+{ x }^{ 2 } } |+C$$
$$\int { \sqrt { 1+{ x }^{ 2 } } }$$dx
= $$\frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log|x+\sqrt { 1+{ x }^{ 2 } } |+C$$

Question 11.
$$\int \sqrt{x^{2}-8 x+7}$$ dx

Solution:

Question 12.
Integrate $$x \sqrt{x+x^{2}}$$ w.r.t. x.
Solution:
I = $$x \sqrt{x+x^{2}}$$ dx
Let x = A$$\frac { d }{ dx }$$(x + x²) + Bx = A(1 + 2x) + B
Equating coefficients of x both sides, we get 2A = 1 ⇒ A = $$\frac { 1 }{ 2 }$$
Equating constants on both sides, we get A + B = 0 ⇒ B = $$\frac { – 1 }{ 2 }$$

Question 13.
Integrate $$(x+1) \sqrt{2 x^{2}+3}$$ w.r.t. x.
Solution:
I = $$(x+1) \sqrt{2 x^{2}+3}$$ dx
Let x + 1 = A$$\frac { d }{ dx }$$(2x² + 3) + Bx = A(4x) + B
Equating the coefficients of x both sides, we get 4A = 1 ⇒ A = $$\frac { 1 }{ 4 }$$
Equating constants on both sides, we get B = 1
x + 1 = $$\frac { 1 }{ 4 }$$(4x) + 1

Question 14.
Integrate $$(x+3) \sqrt{3-4 x-x^{2}}$$ w.r.t. x.
Solution:
I = $$(x+3) \sqrt{3-4 x-x^{2}}$$ dx
Let x + 3 = A$$\frac { d }{ dx }$$(3 – 4x – x²) + Bx + 3 = A(- 4 – 2x) + B
Equating the coefficients of x both sides, we get -2A = 1 ⇒ A = $$\frac { – 1 }{ 2 }$$
Equating constants on both sides, we get -4A + B = 3 ⇒ B = 1
∴ x + 3 = $$\frac { – 1 }{ 2 }$$(- 4 – 2x) + 1

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