Question

the value of $\int \frac{\ell n\left(\frac{x-1}{x+1}\right)}{x^2-1}dxisequalis$

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

The correct option is B $\frac{1}{4}{\ell n}^2\frac{x-1}{x+1}+C$

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

Let $t=\ln \left(\frac{x-1}{x+1}\right)$

$\Rightarrow dt=\frac{1}{\frac{x-1}{x+1}}\frac{d}{dx}\left(\frac{x-1}{x+1}\right)dx$

$\Rightarrow dt=\frac{x+1}{x-1}\frac{(x+1)\frac{d}{dx}(x-1)-(x-1)\frac{d}{dx}(x+1)}{{(x+1)}^2}dx$

$\Rightarrow dt=\frac{x+1}{x-1}\frac{(x+1)-(x-1)}{{(x+1)}^2}dx$

$\Rightarrow dt=\frac{x+1}{x-1}\frac{x+1-x+1}{{(x+1)}^2}dx$

$\Rightarrow dt=\frac{x+1}{x-1}\frac{2}{{(x+1)}^2}dx$

$\Rightarrow \frac{1}{2}dt=\frac{dx}{x-1}\frac{1}{(x+1)}$

$\Rightarrow \frac{1}{2}dt=\frac{dx}{x^2-1}$

$\therefore \int \frac{\ln \left(\frac{x-1}{x+1}\right)}{x^2-1}dx$

$=\frac{1}{2}\int tdt$

$=\frac{1}{2}\frac{t^2}{2}+c$

$=\frac{1}{4}{t}^2+c$

$=\frac{1}{4}{t}^2+c$ where $t=\ln \left(\frac{x-1}{x+1}\right)$

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

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