Anand Classes provides a comprehensive, free-downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 7 – Integrals, Exercise 7.4 (Set-2 Study Material). This study material has been carefully crafted to cover the integrals of some particular functions, including advanced problems with step-by-step solutions aligned to the latest CBSE/NCERT syllabus. Click the print button to download study material and notes.
NCERT Question 13: Evaluate the integral
$$\int \frac{1}{\sqrt{(x-1)(x-2)}}\ dx$$
Solution
$$(x-1)(x-2)=x^{2}-3x+2$$
Complete the square for $x^{2}-3x+2$:
$$x^{2}-3x+2 = x^{2}-3x+\frac{9}{4}-\frac{9}{4}+2 $$
$$x^{2}-3x+2 = \left(x-\frac{3}{2}\right)^{2}-\frac{1}{4}$$
$$x^{2}-3x+2 = \left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}$$
Thus the integral becomes
$$\int \frac{1}{\sqrt{\left(x-\dfrac{3}{2}\right)^{2}-\left(\dfrac{1}{2}\right)^{2}}}\ dx$$
Let
$$t = x-\frac{3}{2}$$
so
$$dt = dx$$
Then
$$\int \frac{1}{\sqrt{t^{2}-\left(\dfrac{1}{2}\right)^{2}}}\ dt
= \ln!\left|t+\sqrt{t^{2}-\left(\frac{1}{2}\right)^{2}}\right| + C$$
Substitute back $t = x-\frac{3}{2}$
$$= \ln\left|x-\frac{3}{2}+\sqrt{x^{2}-3x+2}\right| + C$$
$$\boxed{\ln\left|x-\frac{3}{2}+\sqrt{x^{2}-3x+2}\right| + C}$$
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NCERT Question 14: Evaluate the Integral
$$\int \frac{1}{\sqrt{8 + 3x – x^{2}}}\ dx$$
Solution
$$\int \frac{1}{\sqrt{8 + 3x – x^{2}}}\ dx$$
Rewrite the quadratic:
$$8 + 3x – x^{2}
= 8 – (x^{2} – 3x)
= 8 – \left(x^{2}-3x+\frac{9}{4}-\frac{9}{4}\right)$$
$$= 8 – \left[\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}\right]
= 8 – \left(x-\frac{3}{2}\right)^{2} + \frac{9}{4}$$
Combine constants:
$$8 + \frac{9}{4} = \frac{32}{4} + \frac{9}{4} = \frac{41}{4}$$
Thus,
$$8+3x-x^{2} = \frac{41}{4} – \left(x-\frac{3}{2}\right)^{2}$$
So,
$$\int \frac{1}{\sqrt{8+3x-x^{2}}}\ dx
= \int \frac{1}{\sqrt{\dfrac{41}{4}-\left(x-\dfrac{3}{2}\right)^{2}}}\ dx$$
Let
$$t = x – \frac{3}{2}$$
so
$$dt = dx$$
Then the integral becomes
$$\int \frac{1}{\sqrt{\left(\dfrac{\sqrt{41}}{2}\right)^{2}-t^{2}}}\ dt
= \sin^{-1}\left(\frac{t}{\dfrac{\sqrt{41}}{2}}\right) + C$$
Substitute back:
$$= \sin^{-1}\left(\frac{x-\dfrac{3}{2}}{\dfrac{\sqrt{41}}{2}}\right) + C$$
Multiply numerator and denominator by $2$:
$$= \sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right) + C$$
$$\boxed{\sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right) + C}$$
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NCERT Question 15: Evaluate the integral
$$\int \frac{1}{\sqrt{(x-a)(x-b)}}\ dx$$
Solution
$$\int \frac{1}{\sqrt{(x-a)(x-b)}}\ dx$$
$$(x-a)(x-b)=x^{2}-(a+b)x+ab$$
Complete the square:
$$x^{2}-(a+b)x+ab= x^{2}-(a+b)x+\frac{(a+b)^{2}}{4}-\frac{(a+b)^{2}}{4}+ab$$
$$x^{2}-(a+b)x+ab = \left(x-\frac{a+b}{2}\right)^{2}-\frac{(a+b)^{2}-4ab}{4}$$
$$x^{2}-(a+b)x+ab = \left(x-\frac{a+b}{2}\right)^{2}-\frac{(a-b)^{2}}{4}$$
Thus the integral becomes
$$\int \frac{1}{\sqrt{\left(x-\dfrac{a+b}{2}\right)^{2}-\left(\dfrac{a-b}{2}\right)^{2}}}\ dx$$
Let
$$t = x-\frac{a+b}{2}$$
so
$$dt = dx$$
Then
$$\int \frac{1}{\sqrt{t^{2}-\left(\dfrac{a-b}{2}\right)^{2}}}\ dt
= \ln\left|t+\sqrt{t^{2}-\left(\frac{a-b}{2}\right)^{2}}\right| + C$$
Substitute back $t = x-\dfrac{a+b}{2}$
$$= \ln\left|x-\frac{a+b}{2}+\sqrt{(x-a)(x-b)}\right| + C$$
$$\boxed{\ln\left|x-\frac{a+b}{2}+\sqrt{(x-a)(x-b)}\right| + C}$$
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NCERT Question 16: Evaluate the integral
$$\int \frac{4x + 1}{\sqrt{2x^{2} + x – 3}}\ dx$$
Solution
$$\int \frac{4x + 1}{\sqrt{2x^{2} + x – 3}}\ dx$$
Let $$t = 2x^{2} + x – 3$$
Then $$dt = (4x + 1)\ dx$$
So
$$\int \frac{4x + 1}{\sqrt{2x^{2} + x – 3}}\ dx = \int \frac{dt}{\sqrt{t}}$$
$$= 2\sqrt{t} + C$$
$$= 2\sqrt{2x^{2} + x – 3} + C$$
$$\boxed{2\sqrt{2x^{2} + x – 3} + C,}$$
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NCERT Question 17: Evaluate the integral
$$\int \frac{x+2}{\sqrt{x^{2}-1}}\ dx$$
Solution
$$\int \frac{x+2}{\sqrt{x^{2}-1}}\ dx$$
Write
$$x+2 = A\frac{d}{dx}(x^{2}-1) + B = A(2x) + B$$
Comparing coefficients:
$$2A = 1 \Rightarrow A = \tfrac{1}{2}$$
$$B = 2$$
So
$$\int \frac{x+2}{\sqrt{x^{2}-1}}\ dx = \int \frac{\dfrac{1}{2}(2x) + 2}{\sqrt{x^{2}-1}}\ dx $$
$$ \int \frac{x+2}{\sqrt{x^{2}-1}}\ dx = \dfrac{1}{2}\int \frac{2x}{\sqrt{x^{2}-1}}\ dx + 2\int \frac{1}{\sqrt{x^{2}-1}}\ dx$$
For the first integral let $t = x^{2}-1$ so $dt = 2x\ dx$:
$$\int \frac{x+2}{\sqrt{x^{2}-1}}\ dx= \dfrac{1}{2}\int \frac{2x}{\sqrt{x^{2}-1}}\ dx = \dfrac{1}{2}\int \frac{dt}{\sqrt{t}} $$
$$ \dfrac{1}{2}\int \frac{dt}{\sqrt{t}} = \dfrac{1}{2}\cdot 2\sqrt{t} = \sqrt{t} = \sqrt{x^{2}-1}$$
For the second integral use the standard form:
$$\int \frac{1}{\sqrt{x^{2}-1}}\ dx = \ln\big|x + \sqrt{x^{2}-1}\big| + C$$
Thus
$$2\int \frac{1}{\sqrt{x^{2}-1}}\ dx = 2\ln\big|x + \sqrt{x^{2}-1}\big|$$
Combining both parts:
$$\int \frac{x+2}{\sqrt{x^{2}-1}}\ dx
= \sqrt{x^{2}-1} + 2\ln\big|x + \sqrt{x^{2}-1}\big| + C$$
$$\boxed{\sqrt{x^{2}-1} + 2\ln\big|x + \sqrt{x^{2}-1}\big| + C}$$
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NCERT Question 18: Evaluate the integral
$$\int \frac{5x-2}{1+2x+3x^{2}}\; dx$$
Solution
$$\int \frac{5x-2}{1+2x+3x^{2}}\; dx$$
Write the numerator as a linear combination of the derivative of the denominator and a constant. Let
$$5x-2 = A\frac{d}{dx}\big(1+2x+3x^{2}\big)+B = A(2+6x)+B.$$
Comparing coefficients gives
$$6A = 5\quad\Rightarrow\quad A=\frac{5}{6},$$
$$2A+B=-2\quad\Rightarrow\quad B=-\frac{11}{3}.$$
Thus
$$\frac{5x-2}{1+2x+3x^{2}}=\frac{5}{6}\cdot\frac{2+6x}{1+2x+3x^{2}}-\frac{11}{3}\cdot\frac{1}{1+2x+3x^{2}}.$$
So
$$\int \frac{5x-2}{1+2x+3x^{2}}\; dx
=\frac{5}{6}\int \frac{2+6x}{1+2x+3x^{2}}\; dx-\frac{11}{3}\int \frac{1}{1+2x+3x^{2}}\; dx.$$
Evaluate the first integral. Let
$$t=1+2x+3x^{2}\quad\Rightarrow\quad dt=(2+6x)\; dx.$$
Hence
$$\int \frac{2+6x}{1+2x+3x^{2}}\; dx=\int \frac{dt}{t}=\ln!\big|t\big|+C=\ln!\big|1+2x+3x^{2}\big|+C.$$
Evaluate the second integral by completing the square in the denominator:
$$1+2x+3x^{2}=3\left(x^{2}+\frac{2}{3}x\right)+1
=3\left[\left(x+\frac{1}{3}\right)^{2}+\frac{2}{9}\right].$$
Therefore
$$\int \frac{1}{1+2x+3x^{2}}\; dx
=\int \frac{1}{3\left[\left(x+\dfrac{1}{3}\right)^{2}+\left(\dfrac{\sqrt{2}}{3}\right)^{2}\right]}\; dx $$
$$ \int \frac{1}{1+2x+3x^{2}}\; dx=\frac{1}{3}\int \frac{1}{\left(x+\dfrac{1}{3}\right)^{2}+\left(\dfrac{\sqrt{2}}{3}\right)^{2}}\; dx.$$
Let
$$u=x+\frac{1}{3}\quad\Rightarrow\quad du=dx.$$
So
$$\frac{1}{3}\int \frac{1}{u^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}}\; du
=\frac{1}{3}\cdot\frac{1}{\frac{\sqrt{2}}{3}}\tan^{-1}\Big(\frac{u}{\frac{\sqrt{2}}{3}}\Big)+C $$
$$\frac{1}{3}\int \frac{1}{u^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}}\; du =\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{3x+1}{\sqrt{2}}\Big)+C.$$
Combine the two parts:
$$\int \frac{5x-2}{1+2x+3x^{2}}\; dx
=\frac{5}{6}\ln\big|1+2x+3x^{2}\big|-\frac{11}{3}\cdot\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{3x+1}{\sqrt{2}}\Big)+C$$
$$\boxed{\frac{5}{6}\ln\big|1+2x+3x^{2}\big|-\frac{11}{3\sqrt{2}}\tan^{-1}\Big(\frac{3x+1}{\sqrt{2}}\Big)+C}$$
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NCERT Question 19: Evaluate the integral
$$\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\ dx$$
Solution
$$\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\ dx
= \int \frac{6x+7}{\sqrt{x^{2}-9x+20}}\ dx$$
Write the numerator as a combination of the derivative of the denominator and a constant. Let
$$6x+7 = A\frac{d}{dx}\big(x^{2}-9x+20\big)+B = A(2x-9)+B.$$
Comparing coefficients gives
$$2A = 6 \Rightarrow A = 3,$$
$$-9A + B = 7 \Rightarrow B = 34.$$
Hence
$$\frac{6x+7}{\sqrt{x^{2}-9x+20}}
= 3\cdot\frac{2x-9}{\sqrt{x^{2}-9x+20}} + 34\cdot\frac{1}{\sqrt{x^{2}-9x+20}}.$$
So
$$\int \frac{6x+7}{\sqrt{x^{2}-9x+20}}\ dx
= 3\int \frac{2x-9}{\sqrt{x^{2}-9x+20}}\ dx + 34\int \frac{1}{\sqrt{x^{2}-9x+20}}\ dx$$
For the first integral let
$$t = x^{2}-9x+20\quad\Rightarrow\quad dt = (2x-9)\ dx.$$
Thus
$$\int \frac{2x-9}{\sqrt{x^{2}-9x+20}}\ dx = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t}
= 2\sqrt{x^{2}-9x+20}.$$
For the second integral complete the square:
$$x^{2}-9x+20 = \left(x-\frac{9}{2}\right)^{2}-\frac{1}{4}
= \left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}.$$
Hence
$$\int \frac{1}{\sqrt{x^{2}-9x+20}}\ dx
= \int \frac{1}{\sqrt{\left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}\ dx$$
$$\int \frac{1}{\sqrt{\left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}\ dx = \ln\Big|x-\frac{9}{2}+\sqrt{x^{2}-9x+20}\Big| + C.$$
Combine the two results:
$$\int \frac{6x+7}{\sqrt{x^{2}-9x+20}}\ dx
= 3\cdot 2\sqrt{x^{2}-9x+20} \;+ \\[10pt] + \;34\ln\Big|x-\frac{9}{2}+\sqrt{x^{2}-9x+20}\Big| + C$$
$$\boxed{\;6\sqrt{x^{2}-9x+20}\;+\;34\ln\Big|x-\frac{9}{2}+\sqrt{x^{2}-9x+20}\Big|+C\;}$$
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NCERT Question 20: Evaluate the integral
$$\int \frac{x+2}{\sqrt{4x – x^{2}}}\ dx$$
Solution
$$\int \frac{x+2}{\sqrt{4x – x^{2}}}\ dx$$
Write the numerator as a combination of the derivative of the denominator and a constant
$$x+2 = A\frac{d}{dx}(4x – x^{2}) + B = A(4 – 2x) + B$$
Compare coefficients
$$-2A = 1 \Rightarrow A = -\tfrac{1}{2}$$
$$4A + B = 2 \Rightarrow -2 + B = 2 \Rightarrow B = 4$$
Thus
$$\frac{x+2}{\sqrt{4x – x^{2}}}
= -\dfrac{1}{2}\cdot\frac{4 – 2x}{\sqrt{4x – x^{2}}} + 4\cdot\frac{1}{\sqrt{4x – x^{2}}}$$
So
$$\int \frac{x+2}{\sqrt{4x – x^{2}}}\ dx
= -\dfrac{1}{2}\int \frac{4 – 2x}{\sqrt{4x – x^{2}}}\ dx + 4\int \frac{1}{\sqrt{4x – x^{2}}}\ dx$$
For the first integral let
$$t = 4x – x^{2}\quad\Rightarrow\quad dt = (4 – 2x)\ dx$$
Hence
$$\int \frac{4 – 2x}{\sqrt{4x – x^{2}}}\ dx
= \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{4x – x^{2}}$$
So the first part gives
$$- \dfrac{1}{2}\cdot 2\sqrt{4x – x^{2}} = -\sqrt{4x – x^{2}}$$
For the second integral complete the square in the radicand
$$4x – x^{2} = 2^2 – (x-2)^{2}$$
Thus
$$\int \frac{1}{\sqrt{4x – x^{2}}}\ dx
= \int \frac{1}{\sqrt{2^2 – (x-2)^{2}}}\ dx
= \sin^{-1}\Big(\frac{x-2}{2}\Big) + C$$
Multiply by 4 to get the second part
$$4\sin^{-1}\Big(\frac{x-2}{2}\Big)$$
Combine both parts
$$\int \frac{x+2}{\sqrt{4x – x^{2}}}\ dx
= -\sqrt{4x – x^{2}} + 4\sin^{-1}\Big(\frac{x-2}{2}\Big) + C$$
$$\boxed{-\sqrt{4x – x^{2}} + 4\sin^{-1}\Big(\frac{x-2}{2}\Big) + C}$$
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NCERT Question 21: Evaluate the integral
$$\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\ dx$$
Solution
$$\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\ dx$$
Split and factor:
$$\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\ dx
= \frac{1}{2}\int \frac{2x+4}{\sqrt{x^{2}+2x+3}}\ dx $$
$$ \frac{1}{2}\int \frac{2x+4}{\sqrt{x^{2}+2x+3}}\ dx = \frac{1}{2}\int \frac{2x+2}{\sqrt{x^{2}+2x+3}}\ dx + \int \frac{1}{\sqrt{x^{2}+2x+3}}\ dx $$
$$ \frac{1}{2}\int \frac{2x+2}{\sqrt{x^{2}+2x+3}}\ dx + \int \frac{1}{\sqrt{x^{2}+2x+3}}\ dx = \frac{1}{2} I_1 + I_2$$
where
$$I_1 = \int \frac{2x+2}{\sqrt{x^{2}+2x+3}}\ dx,\quad I_2 = \int \frac{1}{\sqrt{x^{2}+2x+3}}\ dx.$$
Step 1: Evaluate $(I_1)$
Let
$$t = x^{2}+2x+3 \quad \Rightarrow \quad dt = (2x+2)\ dx.$$
Then
$$I_1 = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{x^{2}+2x+3}.$$
Step 2: Evaluate $(I_2)$
Complete the square:
$$x^{2}+2x+3 = (x+1)^{2}+2.$$
Then
$$I_2 = \int \frac{1}{\sqrt{(x+1)^{2}+(\sqrt{2})^{2}}}\ dx
= \ln\big|(x+1)+\sqrt{x^{2}+2x+3}\big|.$$
Step 3: Combine results
$$\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\ dx = \frac{1}{2} I_1 + I_2 $$
$$\frac{1}{2} I_1 + I_2 = \sqrt{x^{2}+2x+3} + \ln\big|(x+1)+\sqrt{x^{2}+2x+3}\big| + C$$
$$\boxed{\sqrt{x^{2}+2x+3} + \ln\big|(x+1)+\sqrt{x^{2}+2x+3}\big| + C}$$
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NCERT Question 22: Evaluate the integral
$$\int \frac{x+3}{x^{2}-2x-5}\; dx$$
Solution
$$\int \frac{x+3}{x^{2}-2x-5}\; dx$$
Write the numerator as a combination of the derivative of the denominator and a constant:
$$(x+3) = A\frac{d}{dx}(x^{2}-2x-5) + B = A(2x-2) + B$$
Compare coefficients:
$$2A = 1 \Rightarrow A = \frac{1}{2}, \quad -2A + B = 3 \Rightarrow B = 4$$
Thus
$$(x+3) = \frac{1}{2}(2x-2) + 4$$
So the integral becomes
$$\int \frac{x+3}{x^{2}-2x-5}\ dx
= \frac{1}{2}\int \frac{2x-2}{x^{2}-2x-5}\ dx + 4\int \frac{1}{x^{2}-2x-5}\ dx $$
$$\frac{1}{2}\int \frac{2x-2}{x^{2}-2x-5}\ dx + 4\int \frac{1}{x^{2}-2x-5}\ dx = \frac{1}{2} I_1 + 4 I_2 $$
Step 1: Evaluate $(I_1)$
Let
$$t = x^{2}-2x-5 \quad \Rightarrow \quad dt = (2x-2)\ dx$$
Then
$$I_1 = \int \frac{2x-2}{x^{2}-2x-5}\ dx = \int \frac{dt}{t} = \ln|t| = \ln|x^{2}-2x-5|$$
Step 2: Evaluate $(I_2)$
Complete the square:
$$x^{2}-2x-5 = (x-1)^{2}-6$$
So
$$I_2 = \int \frac{1}{(x-1)^{2}-(\sqrt{6})^{2}}\ dx
= \frac{1}{2\sqrt{6}} \ln\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|$$
Step 3: Combine results
$$\int \frac{x+3}{x^{2}-2x-5}\ dx
= \frac{1}{2} \ln|x^{2}-2x-5| + 4\cdot \frac{1}{2\sqrt{6}} \ln\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right| + C$$
Simplify:
$$\int \frac{x+3}{x^{2}-2x-5}\ dx
= \frac{1}{2} \ln|x^{2}-2x-5| + \frac{2}{\sqrt{6}} \ln\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right| + C$$
$$\boxed{\frac{1}{2} \ln|x^{2}-2x-5| + \frac{2}{\sqrt{6}} \ln\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right| + C}$$
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NCERT Question 23: Evaluate the integral
$$\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\ dx$$
Solution
$$\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\ dx$$
Write the numerator as a combination of the derivative of the denominator and a constant:
$$(5x+3) = A\frac{d}{dx}(x^{2}+4x+10) + B = A(2x+4) + B$$
Compare coefficients:
$$2A = 5 \Rightarrow A = \frac{5}{2}, \quad 4A + B = 3 \Rightarrow B = -7$$
Thus
$$5x+3 = \frac{5}{2}(2x+4) – 7$$
So the integral becomes
$$\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\ dx
= \frac{5}{2}\int \frac{2x+4}{\sqrt{x^{2}+4x+10}}\ dx – 7\int \frac{1}{\sqrt{x^{2}+4x+10}}\ dx $$
$$\frac{5}{2}\int \frac{2x+4}{\sqrt{x^{2}+4x+10}}\ dx – 7\int \frac{1}{\sqrt{x^{2}+4x+10}}\ dx = \frac{5}{2} I_1 – 7 I_2$$
Step 1: Evaluate $(I_1)$
Let
$$t = x^{2}+4x+10 \quad \Rightarrow \quad dt = (2x+4)\ dx$$
Then
$$I_1 = \int \frac{2x+4}{\sqrt{x^{2}+4x+10}}\ dx = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{x^{2}+4x+10}$$
Step 2: Evaluate $(I_2)$
Complete the square:
$$x^{2}+4x+10 = (x+2)^{2}+6$$
Then
$$I_2 = \int \frac{1}{\sqrt{(x+2)^{2}+6}}\ dx = \ln\big|(x+2)+\sqrt{x^{2}+4x+10}\big|$$
Step 3: Combine results
$$\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\ dx
= \frac{5}{2}\cdot 2\sqrt{x^{2}+4x+10} \;-\\[10pt] -\; 7 \ln\big|(x+2)+\sqrt{x^{2}+4x+10}\big| + C$$
Simplify:
$$\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\ dx
= 5\sqrt{x^{2}+4x+10} \;- \\[10pt] -\; 7 \ln\big|(x+2)+\sqrt{x^{2}+4x+10}\big| + C$$
$$\boxed{\;5\sqrt{x^{2}+4x+10} \;-\; 7 \ln\big|(x+2)+\sqrt{x^{2}+4x+10}\big| + C\;}$$
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NCERT Question 24: Evaluate the integral
$$\int \frac{dx}{x^{2}+2x+2}$$
Options:
(A) $x \tan^{-1}(x+1) + C$
(B) $\tan^{-1}(x+1) + C$
(C) $(x+1) \tan^{-1}x + C$
(D) $\tan^{-1}x + C$
Solution
$$\int \frac{dx}{x^{2}+2x+2}$$
Complete the square in the denominator:
$$x^{2}+2x+2 = (x^{2}+2x+1)+1 = (x+1)^{2}+1$$
Substitute $u = x+1$, so $du = dx$. Then the integral becomes
$$\int \frac{du}{u^{2}+1}$$
This is a standard integral:
$$\int \frac{du}{u^{2}+1} = \tan^{-1}(u) + C$$
Substitute back $u = x+1$:
$$\int \frac{dx}{x^{2}+2x+2} = \tan^{-1}(x+1) + C$$
Correct option: (B) $\tan^{-1}(x+1) + C$
$$\boxed{\tan^{-1}(x+1) + C}$$
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NCERT Question 25: Evaluate the integral
$$\int \frac{dx}{\sqrt{9x-4x^{2}}}$$
Options:
(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) $\frac{1}{2} \sin^{-1}\left(\frac{9x-8}{8}\right) + C$
Solution
Step 1:
Factor out $-4$ from the quadratic under the square root to make it a standard form:
$$9x-4x^2 = -4\left(x^2 – \frac{9}{4}x\right) $$
$$9x-4x^2 = -4\left(x^2 – \frac{9}{4}x + \frac{81}{64} – \frac{81}{64}\right) = -4\left[\left(x-\frac{9}{8}\right)^2 – \left(\frac{9}{8}\right)^2\right]$$
Thus, the integral becomes:
$$\int \frac{dx}{\sqrt{9x-4x^2}} = \int \frac{dx}{\sqrt{-4\left[(x-\frac{9}{8})^2 – (\frac{9}{8})^2\right]}} = \frac{1}{2} \int \frac{dx}{\sqrt{(\frac{9}{8})^2 – (x-\frac{9}{8})^2}}$$
Step 2:
Use the standard formula for the integral
$$\int \frac{dx}{\sqrt{a^2-(x-b)^2}} = \sin^{-1}\left(\frac{x-b}{a}\right) + C$$
Here, $a = \frac{9}{8}$ and $b = \frac{9}{8}$:
$$\frac{1}{2} \int \frac{dx}{\sqrt{(\frac{9}{8})^2 – (x-\frac{9}{8})^2}} = \frac{1}{2} \sin^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right) + C$$
Step 3: Final Answer
$$\int \frac{dx}{\sqrt{9x-4x^2}} = \frac{1}{2} \sin^{-1}\left(\frac{8x-9}{9}\right) + C$$
Correct option: (B)
$$\boxed{\frac{1}{2} \sin^{-1}\left(\frac{8x-9}{9}\right) + C}$$
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