CBSE Class 12 Mathematics Chapter 7 Integrals – NCERT Solutions 2025-26

Exercise 7.7

1. Integrate $\sqrt {4 - {x^2}} $.

Ans: Let $I = \int {\sqrt {4 - {x^2}} dx = \int {\sqrt {{{\left( 2 \right)}^2} - {{\left( x \right)}^2}} dx} } $

It is known that,

$\sqrt {{a^2} - {x^2}} dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + C$ 

For $a$ substitute $2$ and solve the integral.

$\therefore I = \dfrac{x}{2}\sqrt {4 - {x^2}}  + \dfrac{4}{2}{\sin ^{ - 1}}\dfrac{x}{2} + C$ 

$ = \dfrac{x}{2}\sqrt {4 - {x^2}}  + 2{\sin ^{ - 1}}\dfrac{x}{2} + C$

Where $C$ is an arbitrary constant.

The integral of $\sqrt {4 - {x^2}} $ is $\dfrac{x}{2}\sqrt {4 - {x^2}}  + 2{\sin ^{ - 1}}\dfrac{x}{2} + C$.

2. Integrate $\sqrt {1 - 4{x^2}} $.

Ans: Let $I = \int {\sqrt {1 - 4{x^2}} dx = \int {\sqrt {{{\left( 1 \right)}^2} - {{\left( {2x} \right)}^2}} dx} } $

Let $2x = t \Rightarrow 2dx = dt$ 

$\therefore I = \dfrac{1}{2}\int {\sqrt {{{\left( 1 \right)}^2} - {{\left( t \right)}^2}} } $ 

It is known that,

$\sqrt {{a^2} - {x^2}} dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + C$ 

For $a$ substitute $1$ then substitute $2x$ for $t$ and solve the integral.

$\therefore I = \dfrac{1}{2}\left[ {\dfrac{t}{2}\sqrt {1 - {t^2}}  + \dfrac{1}{2}{{\sin }^{ - 1}}t} \right] + C$

$ = \dfrac{t}{4}\sqrt {1 - {t^2}}  + \dfrac{1}{4}{\sin ^{ - 1}}t + C$

$ = \dfrac{{2x}}{4}\sqrt {1 - 4{x^2}}  + \dfrac{1}{4}{\sin ^{ - 1}}2x + C$

$ = \dfrac{x}{2}\sqrt {1 - 4{x^2}}  + \dfrac{1}{4}{\sin ^{ - 1}}2x + C$ 

Where $C$ is an arbitrary constant.

The integral of $\sqrt {1 - 4{x^2}} $ is $\dfrac{x}{2}\sqrt {1 - 4{x^2}}  + \dfrac{1}{4}{\sin ^{ - 1}}2x + C$.

3. Integrate $\sqrt {{x^2} + 4x + 6} $.

Ans: Let $I = \int {\sqrt {{x^2} + 4x + 6} dx} $

$I = \int {\sqrt {{x^2} + 4x + 4 + 2} dx} $

$I = \int {\sqrt {\left( {{x^2} + 4x + 4} \right) + 2} dx} $

$I = \int {\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} dx} $

It is known that,

$\sqrt {{x^2} + {a^2}} dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}}  + \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$ 

For $a$ substitute $\sqrt 2 $ and for $x$ put $x + 2$ and solve the integral.

$\therefore I = \dfrac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x + 6}  + \dfrac{2}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 6} } \right| + C$

$ = \dfrac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x + 6}  + \log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 6} } \right| + C$

Where $C$ is an arbitrary constant.

The integral of $\sqrt {{x^2} + 4x + 6} $ is $\dfrac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x + 6}  + \log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 6} } \right| + C$.

4. Integrate $\sqrt {{x^2} + 4x + 1} $.

Ans: Let $I = \int {\sqrt {{x^2} + 4x + 1} dx} $

$I = \int {\sqrt {\left( {{x^2} + 4x + 4} \right) - 3} dx} $

$I = \int {\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} dx} $

It is known that,

$\sqrt {{x^2} - {a^2}} dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}}  - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$ 

For $a$ substitute $\sqrt 3 $ and for $x$ put $x + 2$ and solve the integral.

$\therefore I = \dfrac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x + 1}  - \dfrac{3}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 1} } \right| + C$

Where $C$ is an arbitrary constant.

The integral of $\sqrt {{x^2} + 4x + 1} $ is $\dfrac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x + 1}  - \dfrac{3}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 1} } \right| + C$.

5. Integrate $\sqrt {1 - 4x - {x^2}} $.

Ans: Let $I = \int {\sqrt {1 - 4x - {x^2}} dx} $

$I = \int {\sqrt {1 - \left( {{x^2} + 4x + 4 - 4} \right)} dx} $

$I = \int {\sqrt {1 + 4 - {{\left( {x + 2} \right)}^2}} dx} $

$I = \int {\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( {x + 2} \right)}^2}} dx} $

It is known that,

$\sqrt {{a^2} - {x^2}} dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + C$ 

For $a$ substitute $\sqrt 5 $ and for $x$ put $x + 2$ and solve the integral.

$\therefore I = \dfrac{{\left( {x + 2} \right)}}{2}\sqrt {1 - 4x - {x^2}}  + \dfrac{5}{2}{\sin ^{ - 1}}\left( {\dfrac{{x + 2}}{{\sqrt 5 }}} \right) + C$

Where $C$ is an arbitrary constant.

The integral of $\sqrt {1 - 4x - {x^2}} $ is $\dfrac{{\left( {x + 2} \right)}}{2}\sqrt {1 - 4x - {x^2}}  + \dfrac{5}{2}{\sin ^{ - 1}}\left( {\dfrac{{x + 2}}{{\sqrt 5 }}} \right) + C$.

6. Integrate $\sqrt {{x^2} + 4x - 5} $.

Ans: Let $I = \int {\sqrt {{x^2} + 4x - 5} dx} $

$I = \int {\sqrt {\left( {{x^2} + 4x + 4} \right) - 9} dx} $

$I = \int {\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( 3 \right)}^2}} dx} $

It is known that,

$\sqrt {{x^2} - {a^2}} dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}}  - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$ 

For $a$ substitute $\sqrt 5 $ and for $x$ put $x + 2$ and solve the integral.

$\therefore I = \dfrac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x - 5}  - \dfrac{9}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x - 5} } \right| + C$

Where $C$ is an arbitrary constant.

The integral of $\sqrt {{x^2} + 4x - 5} $ is $\dfrac{{\left( {x + 2} \right)}}{2}\sqrt {1 - 4x - {x^2}}  + \dfrac{5}{2}{\sin ^{ - 1}}\left( {\dfrac{{x + 2}}{{\sqrt 5 }}} \right) + C$.

7. Integrate $\sqrt {1 + 3x - {x^2}} $.

Ans: Let $I = \int {\sqrt {1 + 3x - {x^2}} dx} $

$I = \int {\sqrt {1 - \left( {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4}} \right)} dx} $

$I = \int {\sqrt {\left( {1 + \dfrac{9}{4}} \right) - {{\left( {x - \dfrac{3}{2}} \right)}^2}} dx} $

$I = \int {\sqrt {{{\left( {\dfrac{{\sqrt {13} }}{2}} \right)}^2} - {{\left( {x - \dfrac{3}{2}} \right)}^2}} dx} $

It is known that,

$\sqrt {{a^2} - {x^2}} dx = \dfrac{x}{2}\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + C$ 

For $a$ substitute $\dfrac{{\sqrt {13} }}{2}$ and for $x$ put $x - \dfrac{3}{2}$ and solve the integral.

$\therefore I = \dfrac{{\left( {x - \dfrac{3}{2}} \right)}}{2}\sqrt {1 - 3x - {x^2}}  + \dfrac{{13}}{{4 \times 2}}{\sin ^{ - 1}}\left( {\dfrac{{x - \dfrac{3}{2}}}{{\dfrac{{\sqrt {13} }}{2}}}} \right) + C$

$ = \dfrac{{2x - 3}}{4}\sqrt {1 - 3x - {x^2}}  + \dfrac{{13}}{8}{\sin ^{ - 1}}\left( {\dfrac{{2x - 3}}{{\sqrt {13} }}} \right) + C$

Where $C$ is an arbitrary constant.

The integral of $\sqrt {1 + 3x - {x^2}} $ is $\dfrac{{2x - 3}}{4}\sqrt {1 - 3x - {x^2}}  + \dfrac{{13}}{8}{\sin ^{ - 1}}\left( {\dfrac{{2x - 3}}{{\sqrt {13} }}} \right) + C$.

8. Integrate $\sqrt {{x^2} + 3x} $.

Ans: Let $I = \int {\sqrt {{x^2} + 3x} dx} $

$I = \int {\sqrt {{x^2} + 3x + \dfrac{9}{4} - \dfrac{9}{4}} dx} $

$I = \int {\sqrt {{{\left( {x + \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{3}{2}} \right)}^2}} dx} $

It is known that,

$\sqrt {{x^2} - {a^2}} dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}}  - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$ 

For $a$ substitute $\dfrac{3}{2}$ and for $x$ put $x + \dfrac{3}{2}$ and solve the integral.

$\therefore I = \dfrac{{x + \dfrac{3}{2}}}{2}\sqrt {{x^2} + 3x}  - \dfrac{{\dfrac{9}{4}}}{2}\log \left| {\left( {x + \dfrac{3}{2}} \right) + \sqrt {{x^2} + 3x} } \right| + C$

$ = \dfrac{{\left( {2x + 3} \right)}}{2}\sqrt {{x^2} + 3x}  - \dfrac{9}{8}\log \left| {\left( {x + \dfrac{3}{2}} \right) + \sqrt {{x^2} + 3x} } \right| + C$

Where $C$ is an arbitrary constant.

The integral of $\sqrt {{x^2} + 3x} $ is $\dfrac{{\left( {2x + 3} \right)}}{2}\sqrt {{x^2} + 3x}  - \dfrac{9}{8}\log \left| {\left( {x + \dfrac{3}{2}} \right) + \sqrt {{x^2} + 3x} } \right| + C$.

9. Integrate $\sqrt {1 + \dfrac{{{x^2}}}{9}} $.

Ans: Let $I = \int {\sqrt {1 + \dfrac{{{x^2}}}{9}} dx} $

$I = \dfrac{1}{3}\int {\sqrt {9 + {x^2}} dx} $

$I = \dfrac{1}{3}\int {\sqrt {{{\left( 3 \right)}^2} + {{\left( x \right)}^2}} dx} $

It is known that,

$\sqrt {{x^2} + {a^2}} dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}}  + \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$ 

For $a$ substitute $3$ and solve the integral.

$\therefore I = \dfrac{1}{3}\left[ {\dfrac{x}{2}\sqrt {{x^2} + 9}  + \dfrac{9}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right|} \right] + C$

$ = \dfrac{x}{6}\sqrt {{x^2} + 9}  + \dfrac{3}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right| + C$

Where $C$ is an arbitrary constant.

The integral of $\sqrt {1 + \dfrac{{{x^2}}}{9}} $ is $\dfrac{x}{6}\sqrt {{x^2} + 9}  + \dfrac{3}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right| + C$.

10. $\int {\sqrt {1 + {x^2}} } $ Is equal to

A. $\dfrac{x}{2}\sqrt {1 + {x^2}}  + \dfrac{1}{2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + C$

B. $\dfrac{2}{3}{\left( {1 + {x^2}} \right)^{\dfrac{2}{3}}} + C$ 

C. $\dfrac{2}{3}x{\left( {1 + {x^2}} \right)^{\dfrac{2}{3}}} + C$

D. $\dfrac{{{x^3}}}{2}\sqrt {1 + {x^2}}  + \dfrac{1}{2}{x^2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + C$ 

Ans: It is known that,

$\sqrt {{x^2} + {a^2}} dx = \dfrac{x}{2}\sqrt {{x^2} + {a^2}}  + \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C$

Let $I = \int {\sqrt {1 + {x^2}} } dx$

For $a$ substitute $1$ and solve the integral.

$\therefore \int {\sqrt {1 + {x^2}} } dx = \dfrac{x}{2}\sqrt {1 + {x^2}}  + \dfrac{1}{2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + C$

Hence, the correct answer is A.

11. $\int {\sqrt {{x^2} - 8x + 7} } $ Is equal to

A. $\dfrac{1}{2}\left( {x - 4} \right)\sqrt {{x^2} - 8x + 7}  + 9\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C$

B. $\dfrac{1}{2}\left( {x + 4} \right)\sqrt {{x^2} - 8x + 7}  + 9\log \left| {x + 4 + \sqrt {{x^2} - 8x + 7} } \right| + C$ 

C. $\dfrac{1}{2}\left( {x - 4} \right)\sqrt {{x^2} - 8x + 7}  - 3\sqrt 2 \log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C$

D. $\dfrac{1}{2}\left( {x - 4} \right)\sqrt {{x^2} - 8x + 7}  - \dfrac{9}{2}\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C$ 

Ans: Let $I = \int {\sqrt {{x^2} - 8x + 7} } dx$

$I = \int {\sqrt {\left( {{x^2} - 8x + 16} \right) - 9} dx} $

$I = \int {\sqrt {{{\left( {x - 4} \right)}^2} - {{\left( 3 \right)}^2}} dx} $

It is known that,

$\sqrt {{x^2} - {a^2}} dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}}  - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$ 

For $a$ substitute $3$ and for $x$ put $x - 4$ and solve the integral.

$\therefore I = \dfrac{{\left( {x - 4} \right)}}{2}\sqrt {{x^2} - 8x + 7}  - \dfrac{9}{2}\log \left| {\left( {x - 4} \right) + \sqrt {{x^2} - 8x + 7} } \right| + C$

Hence, the correct answer is D.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.7

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