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 $a=$

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $-\frac{2}{3}$
• D. None of these

Answer 1

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

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

Let $\sqrt{1-x^3}=t$

$I=\int \frac{1}{x\sqrt{1-x^3}}dx=\int \frac{x^2}{x^3\sqrt{1-x^3}}dx\Rightarrow \frac{1}{2\sqrt{1-x^3}}x(-3x^2)dx=dt\Rightarrow I=\frac{2}{3}\int \frac{1}{t^2-1}dt\Rightarrow \frac{3}{2}I=\int \frac{1}{(t+1)(t-1)}dt=\frac{1}{2}\int \frac{(t+1)-(t-1)}{(t+1)(t-1)}dt\Rightarrow 3I=\int \frac{1}{(t-1)}dt-\int \frac{1}{(t+1)}dt=\ln |t-1|-\ln |t+1|+b$

where $b$ is a constant

$\Rightarrow 3I=\ln \left|\frac{t-1}{t+1}\right|+b$

Converting $t=\sqrt{1-x^3}$, we get

$I=\frac{1}{3}\ln \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|$

Hence, the correct answer is $a=\frac{1}{3}$