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	$\int \frac{1}{x-x^3}dx$	1	$\frac{-2}{a}\sqrt{\frac{a-x}{x}}+C$
B	$\int \frac{1}{\sqrt{a+x}+\sqrt{x}+b}dx$	2	$\frac{1}{2}\log \left|\frac{x^2}{1-x^2}\right|+C$
C	$\int \frac{1}{x\sqrt{ax-x^2}}dx$	3	$-{\left(1+x^{-4}\right)}^{1/4}+C$
D	$\int \frac{1}{x^2{\left(x^4+1\right)}^{3/4}}dx$	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. $\int \frac{1}{x-x^3}dx=\int \frac{1}{x\left(1-x^2\right)}dx$
$=\int \frac{1}{x(1-x)(1+x)}dx.....$(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
$\int \frac{1}{x\left(1-x^2\right)}dx=\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{1-x}dx-\frac{1}{2}\int \frac{1}{1+x}dx$
$=\log |x|-\frac{1}{2}\log |1-x|-\frac{1}{2}\log |1+x|+C$
$=\log |x|-\frac{1}{2}\log \{|1+x||1-x|\}+C$
$=\frac{1}{2}\log |x{|}^2-\frac{1}{2}\log \left|1-x^2\right|+C$
$=\frac{1}{2}\log \left|\frac{x^2}{1-x^2}\right|+C$
B. $\int \frac{1}{\sqrt{x+a}}+\sqrt{x+b}dx$
$=\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}}\times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}dx$
$=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}dx$
$=\frac{1}{a-b}\int \left[(x+a{)}^{1/3}-(x+b{)}^{1/3}\right]dx$
$=\frac{1}{a-b}\left[\frac{(x+a{)}^{(1/2)+1}}{\frac{1}{2}+1}-\frac{(x+b{)}^{(1/2)+1}}{\frac{1}{2}+1}\right]+C$
$=\frac{1}{a-b}\left[\frac{(x+a{)}^{3/2}}{3/2}-\frac{(x+b{)}^{3/2}}{3/2}\right]+C$
$=\frac{2}{3(a-b)}\left[(x+a{)}^{3/2}-(x+b{)}^{3/2}\right]+C$
C. Let $I=\int \frac{1}{x\sqrt{ax-x^2}}dx$
Put $x=\frac{a}{t}\Rightarrow dx=\frac{-adt}{t^2}$
$\therefore l=\int \frac{1}{\frac{a}{t}\sqrt{a\left(\frac{a}{t}\right)-\frac{a^2}{t^2}}}\left(\frac{-a}{t^2}\right)dt=\int \frac{-a}{a\cdot a\sqrt{t-1}}dt$
$=-\frac{1}{a}\int (t-1{)}^{-1/2}dt$
$=-\frac{1}{a}\cdot \frac{(t-1{)}^{(-1/2)+1}}{-(1/2)+1}+C$
$=-\frac{2}{a}\sqrt{t-1}+C=-\frac{2}{a}\sqrt{\frac{a}{x}-1}+C$
$\left(\because x=\frac{a}{t}\Rightarrow t=\frac{a}{x}\right)$
$=-\frac{2}{a}\sqrt{\frac{a-x}{x}}+C$
D. Let $I=\int \frac{1}{x^2{\left(x^4+1\right)}^{3/4}}dx$
$=\int \frac{1}{x^2{\left\{x^4\left(1+\frac{1}{x^4}\right)\right\}}^{3/4}}dx$
$=\int \frac{1}{x^2\cdot x^3{\left(1+\frac{1}{x^4}\right)}^{3/4}}dx=\int \frac{1}{x^5{\left(1+\frac{1}{x^4}\right)}^{3/4}}dx$
Put $1+\frac{1}{x^4}=t\Rightarrow -\frac{4}{x^5}dx=dt\Rightarrow \frac{1}{x^5}dx=-\frac{dt}{4}$
$\therefore I=\frac{1}{-4}\int \frac{1}{t^{3/4}}dt=-\frac{1}{4}\left[\frac{t^{1/4}}{1/4}\right]+C=-{\left(1+x^{-4}\right)}^{1/4}+C$