Question

Let $I=\int \frac{2}{{(2-x)}^2}{\left(\frac{2-x}{2+x}\right)}^{\frac{1}{3}}dx=k{\left(f\left(x\right)\right)}^m+C$, then $4km$ equals $(k,m\in R)$

• A. $3$
• B. $4$
• C. $2$
• D. None of these

Answer 1

The correct option is C $2$

Given $I=\int \frac{2}{{(2-x)}^2}{\left(\frac{2-x}{2+x}\right)}^{\frac{1}{3}}dx=k{\left(f\left(x\right)\right)}^m+C$ ....(1)
Consider, $I=\int \frac{2}{{(2-x)}^2}{\left(\frac{2-x}{2+x}\right)}^{\frac{1}{3}}dx$
Putting $\frac{2-x}{2+x}=t^3$
$\Rightarrow x=2\left(\frac{1-t^3}{1+t^3}\right)$
$\Rightarrow 2-x=\frac{4t^3}{1+t^3}$
$\Rightarrow dx=\frac{-12t^2}{{(1+t^3)}^2}dt$
$\therefore I=-\int \frac{2{(1+t^3)}^{2}t\left(12t^2\right)}{16t^6{(1+t^3)}^2}dt$
$=-\frac{3}{2}\int t^{-3}dt$
$=\frac{3}{4}{\left(\frac{2+x}{2-x}\right)}^{\frac{2}{3}}+C$
Put this value in (1), we get on comparing
$k=\frac{3}{4}$, $m=\frac{2}{3}$
$\therefore \because 4km=4\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)=2$