Question

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

Given $\int \frac{1}{x\sqrt{1-x^3}}dx=a\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+b$ ....(1)
Consdier, $I=\int \frac{1}{x\sqrt{1-x^3}}dx$
Putting $1-x^3=y^2\Rightarrow x^3=1-y^2$

$\Rightarrow -3x^{2}dx=2ydy$

$\Rightarrow x^{2}dx=\frac{-2}{3}ydy$
$\int \frac{1}{x\sqrt{1-x^3}}dx=\int \frac{x^2}{x^3\sqrt{1-x^3}}dx$

$=-\frac{2}{3}\int \frac{1}{1-y^2}dy$
$=\frac{1}{3}\log \left|\frac{y-1}{y+1}\right|+C$
$=\frac{1}{3}\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C$

On comparing with (1)

$a=\frac{1}{3}$