Question

If $f\left(x\right)=\int \frac{dx}{\sqrt[3]{3(x+1{)}^2(x-1{)}^4}}=k\sqrt[3]{3\frac{1+x}{1-x}}+c$, then $k$ is equal to

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{3}{2}$

Given, $\int \frac{dx}{\sqrt[3]{3(x+1{)}^2(x-1{)}^4}}=k\sqrt[3]{3\frac{1+x}{1-x}}+c$
Differentiating both sides w.r.t. $x$, we get
$\frac{1}{\sqrt[3]{3(x+1{)}^2(x-1{)}^4}}=k\frac{d}{dx}{\left(\frac{1+x}{1-x}\right)}^{1/3}$
$\Rightarrow \frac{1}{{(x+1)}^{2/3}{(x-1)}^{4/3}}=k\frac{1}{3}{\left(\frac{1+x}{1-x}\right)}^{-2/3}\frac{(1-x)+(1+x)}{{(1-x)}^2}$
$\Rightarrow \frac{1}{{(x+1)}^{2/3}{(x-1)}^{4/3}}=k\frac{1}{3}{\left(\frac{1+x}{x-1}\right)}^{-2/3}\frac{2}{{(x-1)}^2}$
$\Rightarrow k=\frac{3}{2}$