How to integrate $\int _1^{\infty }\frac{dx}{\left(x^2+1\right)\sqrt{x^2-1}}= \;?$ 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?
 A: $x=\cosh(t)$, 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
$$ \int_{0}^{1}\frac{2v}{1+6v+v^2}\,dv.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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With the sub$\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}
A: $$\int _{\frac{\pi }{2}}^0\left(\frac{\sin t}{1+\sin^2 t}\right)\:dt=\int _{\frac{\pi }{2}}^0\left(\frac{\sin t}{2-\cos^2t}\right)\:dt$$
Let $u=\cos t$, then $du=-\sin t \space dt$.
$$\int _{\frac{\pi }{2}}^0\left(\frac{\sin t}{2-\cos^2t}\right)\:dt=-\int _{0}^1\frac{du}{2-u^2}=-\frac{1}{2}\int _{0}^1\frac{du}{1-(\frac{u}{\sqrt2})^2} \\=-\frac{1}{2\sqrt2}[\ln(u+\sqrt2)-\ln(\sqrt2 -u)]_0^1\\=-\frac{1}{2\sqrt2}[\ln(1+\sqrt2)-\ln(\sqrt2 -1)-\ln(\sqrt2)+\ln(\sqrt2)]\\=-\frac{1}{2\sqrt2}[\ln(1+\sqrt2)-\ln(\sqrt2 -1)]=\frac{1}{\sqrt2}\ln\left(\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}\right)$$
A: Let $x=\tan{u} \rightarrow dx=\sec^2{u}$ $du$
Then integral becomes
$\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}} \frac{du}{\sqrt{\tan^2{u}-1}}=-\coth{\tan{u}}|^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$=-1+\coth{1}=0.313035$
