Integration: $\int_{2}^{\infty} \frac{\sqrt x}{x^2-1} dx$ I am having a problem integrating this term, I am not able to solve it by substitution either.
 A: Let $t=\sqrt{x}$ then
\begin{align*}\int_{2}^{\infty} \frac{\sqrt x}{x^2-1} dx&=
\int_{\sqrt{2}}^{\infty} \frac{t}{t^4-1}\cdot (2t\,dt)\\&=
\int_{\sqrt{2}}^{\infty}\left(
\frac{1}{2(t-1)}-\frac{1}{2(t+1)}+\frac{1}{1+t^2}\right)dt\\
&=\left[F(t)\right]_{\sqrt{2}}^{\infty}
\end{align*}
which is the integral of a rational function. Can you take it from here?
A: Hint: change $x=t^2$ to get:
$$\int_{\sqrt{2}}^{\infty} \frac{2t^2}{t^4-1}dt=2\int_{\sqrt{2}}^{\infty} \frac{t^2-1+1}{(t^2-1)(t^2+1)}dt.$$
A: 
$$\mathscr{\text{Let } t= \sqrt x}$$
  $$\mathscr{\text{.:} dt=\frac {1}{2\sqrt x} dx}$$
  $$\mathscr{\text{.:} dx= 2t dt}$$

$$\int_0^\infty \frac{t}{t^4-1} 2t dt$$
$$=\int_0^\infty \frac {2t^2}{t^4-1} dt$$
$$= \int_0^\infty \frac{2t^2}{(t^2-1)(t^2+1)} dt$$
$$= \int_0^\infty \frac{(t^2-1)+(t^2+1)}{(t^2-1)(t^2+1)} dt$$
$$= \int_0^\infty (\frac{1}{t^2-1} +\frac{1}{t^2+1})dt$$
$$= \int_0^\infty \frac{(t+1)-(t-1)}{2(t-1)(t+1)} dt +\int_0^\infty \frac{1}{t^2+1} dt$$
$$= \frac{1}{2} \int_0^\infty (\frac{1}{t-1} -\frac{1}{t+1})dt +\tan^{-1} x$$
$$= \frac{1}{2} \int_0^\infty \frac{1}{t-1} dt - \frac{1}{2} \int_0^\infty \frac{1}{t+1} dt +\tan^{-1} x$$
$$=[\frac{1}{2} (\ln (|t-1|) -\ln (|t+1|)) +\tan^{-1} t]_0^\infty$$

$$=[\frac{1}{2} \ln (|\frac{\sqrt x -1}{\sqrt x+ 1}|) +\tan^{-1} \sqrt x]_0^\infty$$

$$=(lim_{x\to \infty} \frac{1}{2} \ln( \frac{\sqrt x - 1}{\sqrt x+1}) -\tan^{-1} (\infty))-(\frac{1}{2} \ln (1) - \tan^{-1} (0))$$
$$= (lim_{x\to \infty} \frac{1}{2} \ln (\frac{1-\frac{1}{\sqrt x}}{1+\frac{1}{\sqrt x}}) -\frac {π}{2})- 0) $$
$$= 0-\frac{π}{2} $$
$$=- \frac{π}{2} $$

$$ \mathrm{\int_0^\infty \frac {\sqrt x}{x^2-1} dx = - \frac {π}{2} }$$

$$\frac{}{}$$

Edit : The question was changed from
  $\int_0^\infty \frac {\sqrt x}{x^2-1} dx  \to \int_2^\infty \frac {\sqrt x}{x^2-1} dx $
  $$\text{You can similarly solve for this limit}$$

