Question

If $\int \frac{dx}{x\sqrt{1-x^3}}=aln\left(\frac{\sqrt{1-x^3}+b}{-\sqrt{1-x^3}+1}\right)+k$, then

• A. $b=1,a=1$
• B. $b=1,a=-\frac{1}{3}$
• C. $b=1,a=-\frac{2}{3}$
• D. None of these

Answer 1

The correct option is B $b=1,a=-\frac{1}{3}$

$I=\int \frac{dx}{x\sqrt{1-x^3}}Put1-x^3=t^{2}3x^{2}dx=-2tdtI=\frac{-2}{3}\int \frac{dt}{(1-t^2)}=\frac{-1}{3}\int (\frac{1}{1-t}+\frac{1}{1+t})dt.=\frac{-1}{3}log\frac{1+t}{1-t}+c=\frac{-1}{3}log\frac{1+\sqrt{1-x^3}}{1-\sqrt{1-x^3}}+c.$