Question

If $I=\int \frac{dx}{x\sqrt{1-x^3}},$ then I equals

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

Answer 1

The correct option is A $\frac{1}{3}\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C$

Let $I=\int \frac{dx}{x\sqrt{1-x^3}}=\int \frac{x^{2}dx}{x^3\sqrt{1-x^3}}$
Substitute $1-x^3=t^2\Rightarrow -3x^{2}dx=2tdt$
$I=\int \frac{-\frac{2}{3}t}{(1-t^2)t}dt=-\frac{2}{3}\int \frac{dt}{1-t^2}$
$=-\frac{2}{3}\left(\frac{1}{2}\right)\log \left(\frac{1+t}{1-t}\right)+c$
$=\frac{1}{3}\log \left(\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right)+c$