If $\int \frac{\mathrm{d} x}{x \sqrt{1-x^3}}=\mathrm{k} \log \left(\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right)+\mathrm{c},($ where $\mathrm{c}$ is a constant of integration), then value of $\mathrm{k}$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{I}=\int \frac{\mathrm{d} x}{x \sqrt{1-x^3}} \ & =\int \frac{x^2 \mathrm{~d} x}{x^3 \sqrt{1-x^3}} \mathrm{~d} x \ & 	ext { Put } 1-x^3=\mathrm{t}^2 \ & \Rightarrow-3 x^2 \mathrm{~d} x=2 \mathrm{tdt} \ & \Rightarrow x^2 \mathrm{~d} x=\frac{-2 \mathrm{tdt}}{3} \ & \end{aligned}$ $\begin{aligned} 	herefore \quad \mathrm{I} & =\int \frac{-\left(\frac{2 \mathrm{tdt}}{3}ight)}{\left(1-\mathrm{t}^2ight) \mathrm{t}} \ & =-\frac{2}{3} \int \frac{1}{1-\mathrm{t}^2} \mathrm{dt} \ & =\frac{2}{3} \int \frac{1}{\mathrm{t}^2-1} \mathrm{dt} \ & =\frac{2}{3} 	imes \frac{1}{2} \cdot \log \left|\frac{\mathrm{t}-1}{\mathrm{t}+1}ight|+\mathrm{c} \ & =\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}ight|+\mathrm{c} \ 	herefore \quad \mathrm{k} & =\frac{1}{3}\end{aligned}$

If $\int \frac{\mathrm{d} x}{x \sqrt{1-x^3}}=\mathrm{k} \log…

Solution