If $\int \frac{1}{x \sqrt{1-x^3}} d x=a \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+b$ then a is equal to
Answer & explanation
Correct answer: option 1
Multiplying above and below by x2 and put
$1-x^3=t^2$ ∴ $-3 x^2 dx=2 t~dt$
∴ $I=\frac{2}{3} \int \frac{d t}{t^2-1}=\frac{2}{3} . \frac{1}{2} \log \frac{t-1}{t+1}+c$
$I=\frac{1}{3} \log \frac{t-1}{t+1}+c$ ∴ $a=\frac{1}{3}$
Hence (2) is the correct answer.