Question

The value of $\int \frac{dx}{x\sqrt{1-x^3}}$ is equal to

• A. $\frac{1}{3}\ell n\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C$
• B. $\frac{1}{3}\ell n\left|\frac{\sqrt{1-x^2}+1}{\sqrt{1-x^2}-1}\right|+C$
• C. $\frac{1}{3}\ell n\left|\frac{1}{\sqrt{1-x^3}}\right|+C$
• D. $\frac{1}{3}\ell n\left|1-x^3\right|+C$

Answer 1

Answer: (A)

Put $1-x^3=t^2\Rightarrow -3x^{2}dx=2tdt$
$I=\int \frac{dx}{x\sqrt{1-x^3}}=\int \frac{x^{2}dx}{x^3\sqrt{1-x^3}}$
$=-\frac{1}{3}\int \frac{2tdt}{\left(1-t^2\right)\cdot t}=-\frac{2}{3}\int \frac{dt}{1-t^2}=\frac{2}{3}\int \frac{dt}{t^2-1}$
$=\frac{2.1}{3.2}\ell n\left|\frac{1-t}{1+t}\right|=\frac{1}{3}\ell n\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3+1}}\right|+c$