Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$ Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$
Using the third substitution of Euler,
$$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$
we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\sqrt 2 t+1}dt+\frac{1}{2}\int\frac{1+\sqrt 2t}{t^2+\sqrt 2t+1}dt$$
Using substitutions $$u=t^2-\sqrt 2t+1$$ and $$v=t^2+\sqrt 2t+1$$
we get $$\int\frac{1-t^2}{1+t^4}dt=-\frac{\sqrt 2}{4}\ln|u|+\frac{\sqrt 2}{4}\ln|v|=-\frac{\sqrt 2}{4}\ln|t^2-\sqrt 2t+1|+\frac{\sqrt 2}{4}\ln|t^2+\sqrt 2t+1|$$
From $$x=\frac{1+t^2}{1-t^2}\Rightarrow t=\sqrt{\frac{x-1}{x+1}}\Rightarrow$$
$$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx=$$$$-\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}-\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}+\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+c$$
Is there another, quicker method to solve this integral, rather than Euler substitution.
 A: Setting $$x=\sinh(t)$$ then we have  $$dx=\cosh(t)dt$$ and our integral will be $$\int \frac{dt}{1+\sinh^2(t)}$$
then use $$\sinh(t)=-2\,{\frac {\tanh \left( t/2 \right) }{ \left( \tanh \left( t/2
 \right)  \right) ^{2}-1}}
$$
note that $$x^2-1=\sinh^2(t)-1=\cosh^2(t)$$
A: Here's what I tried. Setting $x=\sec \theta$, we get
\begin{align*}
\int \frac{1}{(x^2+1)\sqrt{x^2-1}} \mathrm{d}x &= \int \frac{\sec \theta \tan \theta}{(\sec^2 \theta+1)\sqrt{\sec^2 \theta-1}} \mathrm{d}\theta \\
&= \int \frac{\sec \theta}{\sec^2 \theta+1} \mathrm{d}\theta \\
&= \int \frac{\cos \theta}{\cos^2 \theta+1} \mathrm{d}\theta \\
&= \int \frac{1}{2-\sin^2 \theta} \mathrm{d}(\sin \theta)\\
&= \frac{1}{2\sqrt 2} \ln \left|\frac{\sqrt 2 + \sin \theta}{\sqrt 2 - \sin \theta}\right|+C
\end{align*}
Since, $\sec \theta = x \implies \sin \theta = \frac{\sqrt {x^2-1}}{x}$, we get
$$\int \frac{1}{(x^2+1)\sqrt{x^2-1}} \mathrm{d}x = \frac{1}{2\sqrt 2} \ln \left|\frac{\sqrt 2x + \sqrt{x^2-1}}{\sqrt 2x - \sqrt{x^2-1}}\right| +C$$
A: $$ \begin{align}\int \frac{1}{(1+x^2)\sqrt{x^2 - 1}} \, dx  &= \int \dfrac{\sinh u}{(\cosh^2 u + 1)\sqrt{\cosh^2 u - 1}} \, du && x = \cosh u \\ &= \int \dfrac{\sinh u}{(\cosh^2 u + 1)\sinh u} \, du \\ &= \int \dfrac{1}{\cosh^2 u + 1}\, du \\ && \end{align}$$
After which things become simple in terms of hyperbolic tsngent of u and it's inverse. Unsubstitute after that.
A: $I=\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$
Put $x=\frac{1}{t}\implies dx=-\frac{1}{t^2}dt \implies
I=-\int\frac{t}{(t^2+1)\sqrt{1-t^2}}dt  
$  Put$ 1-t^2=y^2\implies -t dt=y dy. \implies
I=\int\frac{y}{(2-y^2)y}dy
=\frac{1}{2\sqrt2}ln\frac{\sqrt2+y}{\sqrt2-y}+C
=\frac{1}{2\sqrt2}ln\frac{\sqrt2+\sqrt{1-t^2}}{\sqrt2-\sqrt{1-t^2}}+C
=\frac{1}{2\sqrt2}ln\frac{\sqrt2x+\sqrt{x^2-1}}{\sqrt2x-\sqrt{x^2-1}}+C
$
