Question

If $\int \frac{dx}{x\sqrt{1-x^3}}=alog\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C$ then a =

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $-\frac{1}{3}$
• D. $-\frac{2}{3}$

Answer 1

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

Put $1-x^3=t^2\Rightarrow -3x^{2}dx=2tdt\therefore \int \frac{dx}{x\sqrt{1-x^3}}=\int \frac{x^2}{x^3\sqrt{1-x^3}}dx=\frac{2}{3}\int \frac{dt}{t^2-1}=\frac{1}{3}log\left|\frac{t-1}{t+1}\right|+C=\frac{1}{3}log\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C\therefore a=\frac{1}{3}$