Question

$I=\int \frac{\ln \left(\frac{x-1}{x+1}\right)}{x^2-1}dx$ is equal to

• A. $\frac{1}{2}{\left(\ln \left(\frac{x-1}{x+1}\right)\right)}^2+C$
• B. $\frac{1}{2}{\left(\ln \left(\frac{x+1}{x-1}\right)\right)}^2+C$
• C. $\frac{1}{4}{\left(\ln \left(\frac{x-1}{x+1}\right)\right)}^2+C$
• D. $\frac{1}{4}\left(\ln \left(\frac{x+1}{x-1}\right)\right)$

Answer 1

Answer: (C)

$\frac{1}{4}{\left(\ln \left(\frac{x-1}{x+1}\right)\right)}^2+C$

$I=\int \frac{\ln \left(\frac{x-1}{x+1}\right)}{x^2-1}dx$
Let $t=\ln \left(\frac{x-1}{x+1}\right)$
$\Rightarrow \frac{dt}{dx}=\frac{x+1}{x-1}\left\{\frac{x+1-(x-1)}{(x+1{)}^2}\right\}=\frac{2}{(x^2-1)}$
$\Rightarrow \frac{dx}{x^2-1}=\frac{dt}{2}$
$\therefore I=\frac{1}{2}\int tdt$
$=\frac{1}{4}{t}^2+C$
$=\frac{1}{4}{\left(\ln \left(\frac{x-1}{x+1}\right)\right)}^2+C$