Help solving the improper integral $\int_1^\infty \frac{2}{4{x^2}-1}\,\mathrm dx$

Question

I need to solve an improper integral which is:

$$ \int_1^\infty \frac{2}{4{x^2}-1}\,\mathrm dx $$

i was trying to solve it using simple substitution but cannot seem to figure it out, i tried a website to solve the integral and it leads to the result being $$ \frac{i\pi}2 - (-\operatorname{arctanh}(2)) $$

and then it says simplify and it gets $ \frac{\ln(3)}2 $ which is the result i need but i don't get how to get there, also i have never used hyperbolic functions nor imaginary numbers so i was wondering if there's another way to integrate this function without using them (hyperbolic functions and imaginary numbers), if there's no other way could you explain to me the steps?.

This is the link to the site solving the integral : solution

Answer 1

$$ \int_{1}^{\infty}\frac{2}{4x^2-1}dx=\int_{1}^{\infty}\left(\frac{1}{2x-1}-\frac{1}{2x+1}\right)dx=\frac{1}{2}\ln\left| \frac{2x-1}{2x+1}\right|_{1}^{\infty}=\frac{1}{2}\bigl(\lim_{x\to \infty}\ln\left| \frac{2x-1}{2x+1}\right|-\ln\frac{1}{3}\bigr)=\frac{\ln 3}{2}. $$

Answer 2

By setting $x=\frac{1}{y}$, $$\int_{1}^{+\infty}\frac{2}{4x^2-1}\,dx = \int_{0}^{1}\frac{2\,dy}{4-y^2}=\frac{1}{2}\left(\int_{0}^{1}\frac{dy}{2-y}+\frac{dy}{2+y}\right)=\frac{1}{2}\int_{1}^{3}\frac{dz}{z}=\color{red}{\frac{\log 3}{2}}.$$

Answer 3

SMALL HINT

factor out $\frac{1}{2}$ and use this method to make the appropriate sub.