Question

$\int \frac{1}{x\sqrt{1-x^3}}dx=a\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+b$, then a is equal to

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

Answer 1

The correct option is A $\frac{1}{3}$

Let $I=\int \frac{1}{x\sqrt{1-x^3}}dx$
Putting $1-x^3=y^2,-3x^{2}dx=2ydy$, we get
$=-\frac{2}{3}\int \frac{1}{1-y^2}dy$
$=\frac{1}{3}\int [\frac{1}{y-1}-\frac{1}{1+y}]$
$=\frac{1}{3}\log \left|\frac{y-1}{y+1}\right|+C$
$=\frac{1}{3}\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C$
On comparing with given equation, we get $a=\frac{1}{3}$