Question

Match the following integrals in column I with their corresponding values in column II and choose the correct option from the codes given below.

Column I		Column II
A	$\int \frac{1}{x-x^3} d x$	1	$\frac{-2}{a} \sqrt{\frac{a-x}{x}}+C$
B	$\int \frac{1}{\sqrt{a+x}+\sqrt{x}+b} d x$	2	$\frac{1}{2} \log \left|\frac{x^2}{1-x^2}\right|+C$
C	$\int \frac{1}{x \sqrt{a x-x^2}} d x$	3	$-\left(1+x^{-4}\right)^{1 / 4}+C$
D	$\int \frac{1}{x^2\left(x^4+1\right)^{3 / 4}} d x$	4	$\frac{2}{3(a-b)}\left[(x+a)^{3 / 2} -(x+b)^{3 / 2}\right]+C$

• A. A - (1), B - (4), C - (2), D - (3)
• B. A - (2), B - (1), C - (3), D - (4)
• C. A - (2), B - (4), C - (1), D - (3)
• D. A - (1), B - (2), C - (4), D - (3)

Answer 1

Answer: (C)

A. $\int \frac{1}{x-x^3} d x=\int \frac{1}{x\left(1-x^2\right)} d x$
$=\int \frac{1}{x(1-x)(1+x)} d x .....$(i)
Let $\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{(1+x)} $
$ \Rightarrow 1=A(1-x)(1+x)+B(x)(1+x)+C(x)(1-x)$
$ \Rightarrow 1=A\left(1-x^2\right)+B\left(x+x^2\right)+C\left(x-x^2\right) $
$ \Rightarrow 1=x^2(B-A-C)+x(B+C)+A$
On equating the coefficients of $x^2, x$ and constant term on both sides, we get
$-A+B-C=0, B+C=0 \text { and } A=1$
On solving these equations, we get
$A=1, B=\frac{1}{2} \text { and } C=-\frac{1}{2}$
From Eq. (i), we get
$\int \frac{1}{x\left(1-x^2\right)} d x =\int \frac{1}{x} d x+\frac{1}{2} \int \frac{1}{1-x} d x-\frac{1}{2} \int \frac{1}{1+x} d x $
$=\log |x|-\frac{1}{2} \log |1-x|-\frac{1}{2} \log |1+x|+C$
$ =\log |x|-\frac{1}{2} \log \{|1+x||1-x|\}+C $
$=\frac{1}{2} \log |x|^2-\frac{1}{2} \log \left|1-x^2\right|+C $
$ =\frac{1}{2} \log \left|\frac{x^2}{1-x^2}\right|+C$
B. $ \int \frac{1}{\sqrt{x+a}}+\sqrt{x+b} d x $
$=\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}} d x$
$=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} d x $
$=\frac{1}{a-b} \int\left[(x+a)^{1 / 3}-(x+b)^{1 / 3}\right] d x$
$=\frac{1}{a-b}\left[\frac{(x+a)^{(1 / 2)+1}}{\frac{1}{2}+1}-\frac{(x+b)^{(1 / 2)+1}}{\frac{1}{2}+1}\right]+C$
$=\frac{1}{a-b}\left[\frac{(x+a)^{3 / 2}}{3 / 2}-\frac{(x+b)^{3 / 2}}{3 / 2}\right]+C$
$=\frac{2}{3(a-b)}\left[(x+a)^{3 / 2}-(x+b)^{3 / 2}\right]+C$
C. Let $I=\int \frac{1}{x \sqrt{a x-x^2}} d x$
Put $x =\frac{a}{t} \Rightarrow d x=\frac{-a d t}{t^2} $
$\therefore l =\int \frac{1}{\frac{a}{t} \sqrt{a\left(\frac{a}{t}\right)-\frac{a^2}{t^2}}}\left(\frac{-a}{t^2}\right) d t=\int \frac{-a}{a \cdot a \sqrt{t-1}} d t $
$=-\frac{1}{a} \int(t-1)^{-1 / 2} d t$
$ =-\frac{1}{a} \cdot \frac{(t-1)^{(-1 / 2)+1}}{-(1 / 2)+1}+C $
$ =-\frac{2}{a} \sqrt{t-1}+C=-\frac{2}{a} \sqrt{\frac{a}{x}-1}+C$
$ \left(\because x=\frac{a}{t} \Rightarrow t=\frac{a}{x}\right)$
$ =-\frac{2}{a} \sqrt{\frac{a-x}{x}}+C$
D. Let $I=\int \frac{1}{x^2\left(x^4+1\right)^{3 / 4}} d x$
$=\int \frac{1}{x^2\left\{x^4\left(1+\frac{1}{x^4}\right)\right\}^{3 / 4}} d x$
$=\int \frac{1}{x^2 \cdot x^3\left(1+\frac{1}{x^4}\right)^{3 / 4}} d x=\int \frac{1}{x^5\left(1+\frac{1}{x^4}\right)^{3 / 4}} d x$
Put $1+\frac{1}{x^4}=t \Rightarrow-\frac{4}{x^5} d x=d t \Rightarrow \frac{1}{x^5} d x=-\frac{d t}{4}$
$\therefore I=\frac{1}{-4} \int \frac{1}{t^{3 / 4}} d t=-\frac{1}{4}\left[\frac{t^{1 / 4}}{1 / 4}\right]+C=-\left(1+x^{-4}\right)^{1 / 4}+C$