Question

If $I=\int \frac{dx}{\sqrt[3]{3{(x+1)}^2{(x-1)}^4}}$
$=K\sqrt[3]{3\frac{1+x}{1-x}}+C$
then K is equal to

• A. $2/3$
• B. $3/2$
• C. $1/3$
• D. $1/2$

Answer 1

Answer: (B)

$3/2$

Let $I=\int \frac{dx}{\sqrt[3]{3{(x+1)}^2{(x-1)}^4}}=\int \frac{dx}{{(x-1)}^2\sqrt[3]{3{\left(\frac{x+1}{x-1}\right)}^2}}$
Substitute $\frac{x+1}{x-1}=t\Rightarrow \frac{-2}{{(x-1)}^2}dx=dt$
$I=-\frac{1}{2}\int \frac{dt}{\sqrt[3]{3t^2}}=\frac{3}{2}{t}^{1/3}=\frac{3}{2}\sqrt[3]{3\frac{1+x}{1-x}}$

Thus comparing the above expression with the one given in the question, we have, $K=\frac{3}{2}$