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

Answer 1

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

Taking $x^3=si{n}^2\theta$

We get

$I=\int \frac{2cos\theta }{3\sqrt[3]{3sin\theta }\sqrt[3]{3si{n}^2\theta }cos\theta }d\theta =\int \frac{2cosec\theta }{3}d\theta =\frac{2}{3}log|cosec\theta -cot\theta |+C=\frac{2}{3}log|\frac{1-\sqrt{1-x^3}}{\sqrt{x^3}}|+C$

$=\frac{1}{3}log|\frac{(\sqrt{1-x^3}-1{)}^2}{(\sqrt{1-x^3}+1)(\sqrt{1-x^3}-1)}|+C$

$=\frac{1}{3}log|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}|$

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