How to integrate $\int _1^{\infty }\frac{dx}{\left(x^2+1\right)\sqrt{x^2-1}}= \;?$

Question

How do I integrate $\int _1^{\infty }\left(\frac{1}{\left(x^2+1\right)\sqrt{x^2-1}}\right)\:dx$?

So what I've tried is substituting $x\:=\:\frac{1}{\sin t}$. So then I'll have that when $x\rightarrow 1\:t\rightarrow \frac{\pi }{2}$ and when $x\rightarrow \infty \:t\rightarrow 0$.

This would make my integral, after simplifying what I could:$\int _{\frac{\pi }{2}}^0\left(\frac{\sin t}{1+\sin^2t}\right)\:dt$ if I didn't make any mistakes so far.

How do I proceed from here on out?

Answer 1

$x=\cosh(t)$, followed by $t=\log u$, looks as a good substitution. It leads to:

$$ I=\int_{0}^{+\infty}\frac{dt}{1+\cosh^2(t)}=\int_{1}^{+\infty}\frac{4u\,du}{1+6u^2+u^4}=\int_{1}^{+\infty}\frac{2\,dv}{1+6v+v^2}=\color{red}{\frac{\log(3+2\sqrt{2})}{2\sqrt{2}}}$$ by partial fraction decomposition. The RHS can also be written as $\color{red}{\frac{\text{arcsinh}(1)}{\sqrt{2}}}$.

By your way, $\int_{0}^{\pi/2}\frac{\sin t}{1+\sin^2 t}\,dt$ can be tackled though the substitution $t=2\arctan\frac{v}{2}$, leading to $$ \int_{0}^{1}\frac{2v}{1+6v+v^2}\,dv.$$

Answer 2

$$\int _{\frac{\pi }{2}}^0\left(\frac{\sin t}{1+\sin^2 t}\right)\:dt=\int _{\frac{\pi }{2}}^0\left(\frac{\sin t}{2-\cos^2t}\right)\:dt$$

Let $u=\cos t$, then $du=-\sin t \space dt$.

$$\int _{\frac{\pi }{2}}^0\left(\frac{\sin t}{2-\cos^2t}\right)\:dt=-\int _{0}^1\frac{du}{2-u^2}=-\frac{1}{2}\int _{0}^1\frac{du}{1-(\frac{u}{\sqrt2})^2} \\=-\frac{1}{2\sqrt2}[\ln(u+\sqrt2)-\ln(\sqrt2 -u)]_0^1\\=-\frac{1}{2\sqrt2}[\ln(1+\sqrt2)-\ln(\sqrt2 -1)-\ln(\sqrt2)+\ln(\sqrt2)]\\=-\frac{1}{2\sqrt2}[\ln(1+\sqrt2)-\ln(\sqrt2 -1)]=\frac{1}{\sqrt2}\ln\left(\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}\right)$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With the sub$\ds{\ldots\ x \equiv {1 \over \root{1 - t^{2}}}\quad\imp\quad t = {\root{x^{2} - 1} \over x}}$: $$ \left\lbrace\begin{array}{lcl} \ds{\totald{x}{t}} & \ds{=} & \ds{{t \over \pars{1 - t^{2}}^{3/2}}} \\[2mm] \ds{1 \over x^2 + 1} & \ds{=} & \ds{1 - t^{2} \over 2 - t^{2}} \\[2mm] \ds{1 \over \root{x^2 - 1}} & \ds{=} & \ds{\root{1 - t^{2}} \over t} \end{array}\right. $$

---

\begin{align} &\color{#f00}{\int _{1}^{\infty }{\dd x \over \pars{x^{2} + 1}\root{x^{2} - 1}}} = \int _{0}^{1}{\dd t \over 2 - t^{2}} \\[3mm] = &\ {\root{2} \over 4} \int_{0}^{1}\pars{{1 \over \root{2} - t} + {1 \over \root{2} + t}}\,\dd t = {\root{2} \over 4}\,\ln\pars{\root{2} + 1\over \root{2} - 1} \\[3mm] = &\ \color{#f00}{{\root{2} \over 4}\,\ln\pars{3 + 2\root{2}}} \end{align}

Answer 4

Let $x=\tan{u} \rightarrow dx=\sec^2{u}$ $du$

Then integral becomes

$\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}} \frac{du}{\sqrt{\tan^2{u}-1}}=-\coth{\tan{u}}|^{\frac{\pi}{2}}_{\frac{\pi}{4}}$

$=-1+\coth{1}=0.313035$