Question

The value of $\int \frac{\ln \left(\frac{x-1}{x+1}\right)}{x^2-1} d x$ is equal to

• A. $\frac{1}{2} \ell n^2 \frac{x-1}{x+1}+C$
• B. $\frac{1}{4} \ln ^2 \frac{x-1}{x+1}+C$
• C. $\frac{1}{2} \ell n^2 \frac{x+1}{x-1}+C$
• D. $\frac{1}{4} \ell n^2 \frac{x+1}{x-1}+C$

Answer 1

Answer: (D)

$I=\int \frac{\ln \left(\frac{x-1}{x+1}\right)}{x^2-1} d x $
put $\ln \left(\frac{x-1}{x+1}\right)=t \Rightarrow \frac{2}{x^2-1} d x=d t$
$\Rightarrow I=\int t \frac{d t}{2}=\frac{t^2}{4}+C=\log ^2\left(\frac{x-1}{x+1}\right)+C$
$=\frac{1}{4} \log ^2\left(\frac{x+1}{x-1}\right)+C$