Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$ 
Evaluate $$\large{\int} \small{\sqrt{\frac{1+x^2}{x^2-x^4}} \space {\large{dx}}}$$

Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.
 A: In order to evaluate the integral over all domain $ x\in (-1,0)\cup(0,1)$, substitute $t=x\sqrt{x^2}$. Then, $dt =2\sqrt{x^2}\ dx$ and
\begin{align}
\int \sqrt{\frac{1+x^2}{x^2-x^4}}\ dx
 =& \int \frac{1+x^2}{\sqrt{x^2(1-x^4)}}\ dx
=\frac12\int \frac{1+\sqrt{t^2}}{\sqrt{t^2(1-t^2)}}\ dt \\
=& \ \frac12\int \frac{1}{\sqrt{1-t^2}}+ \frac{1}{\sqrt{t^2(1-t^2)}}\ dt\\
=& \ \frac12 \sin^{-1}t-\frac12\sinh^{-1}
\frac{\sqrt{1-t^2}}t \\
= & \ \frac12 \sin^{-1}\left( x\sqrt{x^2}\right)-\frac12\sinh^{-1}\bigg(
\frac1x \sqrt{\frac{1-x^4}{x^2}}\bigg)\\ 
\end{align}
A: Given integral $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Let's factor out $x^2 $ from the denominator.
$\Rightarrow \int \frac{1}{x}\sqrt{\frac{1+x^2}{1-x^2}}dx$
Now multiply and divide the numerator and the denominator by $\sqrt{1+x^2}$
$\Rightarrow \int \frac{1+x^2}{x\sqrt{1-x^4}}dx$
Now we can split the integral into 2 sub-integrals
$\Rightarrow \int \frac{1}{x\sqrt{1-x^4}}dx+\int \frac{x^2}{x\sqrt{1-x^4}}dx$
$\Rightarrow \int \frac{x^3}{x^4\sqrt{1-x^4}}dx+\int \frac{x}{\sqrt{1-x^4}}dx$
For the first integral, use $1-x^4=u^2$ and for the second, use $x^2=v$. The integrals reduce to
$\Rightarrow \frac{-1}{2}\int \frac{du}{1-u^2}+\frac{1}{2}\int \frac{dv}{\sqrt{1-v^2}}$
$\Rightarrow \frac{1}{2}\int \frac{du}{u^2-1}+\frac{1}{2} \sin ^{-1}v$
$\Rightarrow \frac{1}{4}\log |\frac{u-1}{u+1}| +\frac{1}{2}\sin ^{-1}v+c$
$\Rightarrow \frac{1}{4}\log |\frac{\sqrt{1-x^4}-1}{\sqrt{1-x^4}+1}| +\frac{1}{2}\sin ^{-1}(x^2)+c$
Hence, $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx=\frac{1}{4}\log |\frac{\sqrt{1-x^4}-1}{\sqrt{1-x^4}+1}| +\frac{1}{2}\sin ^{-1}(x^2)+c$
A: Recall the half-angle identity
$$\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}. \tag{1}$$  This suggests the substitution $$x^2 = \cos 2\theta, \quad x \, dx = -\sin 2\theta \, d\theta$$ which yields for $0 < x < 1$
$$\begin{align}
\int \sqrt{\frac{1+x^2}{x^2-x^4}} \, dx
&= \int \frac{1}{x^2} \sqrt{\frac{1+x^2}{1-x^2}} \cdot  x \, dx  \\
&= \int \sec 2\theta \cot \theta \,(-\sin 2\theta) \, d\theta \\
&= - \int \tan 2\theta \cot \theta \, d\theta \\
&= - \int (1 + \sec 2\theta) \, d\theta \\
&= - \theta - \frac{1}{2} \log (\sec 2\theta + \tan 2\theta) + C \\
&= - \frac{1}{2} \arccos x^2 - \frac{1}{2} \log \left( \frac{1 + \sqrt{1-x^4}}{x^2} \right) + C.
\end{align} $$
A: Substituting $u=x^2,dx=\frac1{2\sqrt u}\,du$ gives
$$\int\sqrt{\frac{1+x^2}{x^2-x^4}}\,dx=\int\frac1x\sqrt{\frac{1+x^2}{1-x^2}}\,dx$$
$$=\int\frac1{2u}\sqrt{\frac{1+u}{1-u}}\,du=\frac12\int\frac{1+u}{u\sqrt{1-u^2}}\,du$$
This breaks into two parts, the first evaluable by the same substitution $v=u^2$ followed by a trigonometric integral and the second just a trigonometric integral:
$$=\frac12\left(\int\frac1{u\sqrt{1-u^2}}\,du+\int\frac1{\sqrt{1-u^2}}\,du\right)$$
$$=\frac12\left(\int\frac1{2v\sqrt{1-v}}\,dv+\sin^{-1}u\right)+K$$
$$=\frac12\left(-\tanh^{-1}\sqrt{1-v}+\sin^{-1}u\right)+K$$
$$=\frac12(\sin^{-1}x^2-\tanh^{-1}\sqrt{1-x^4})+K$$
Note that we assumed $x>0$ in the first step, so the above solution is only valid on that domain. For $x<0$ we need to negate, so a solution valid over $\mathbb R\setminus\{0\}$ is
$$\frac{\operatorname{sgn}(x)}2(\sin^{-1}x^2-\tanh^{-1}\sqrt{1-x^4})+K$$
A: $$\large{\int} \normalsize{\dfrac{1}{x}\cdot\sqrt{\dfrac{1+x^2}{1-x^2}}} \space \mathrm{dx} \quad \xrightarrow{\large{x \space = \space \tan{u}}} \quad \large{\int} \normalsize{\dfrac{\cos{u}}{\sin{u}}\cdot\sqrt{\dfrac{1+\tan^2{u}}{1-\tan^2{u}}}} \cdot \sec^2{u} \space \mathrm{du} \\ \quad \\$$
$$\require{\cancel} = \large{\int} \normalsize{\dfrac{\cancel{\cos^2{u}}}{\sin{u}}\cdot \dfrac{1}{\cos^\cancel{3}{u}\sqrt{2\cos^2{u} - 1}}} \space \mathrm{du}$$
$$\require{\cancel} = \dfrac{1}{2}\large{\int} \normalsize{\dfrac{1}{\sin{2u} \sqrt{\cos{2u}}}} \space \mathrm{du}$$
$$\require{\cancel} = \dfrac{1}{2}\large{\int} \normalsize{\dfrac{\sin{2u}}{\left(1 - \cos^2{2u}\right) \sqrt{\cos{2u}}}} \space \mathrm{du} \quad \xrightarrow{\large{t \space = \space \cos{2u}}} \quad  \dfrac{1}{2}\large{\int} \normalsize{\dfrac{1}{\sqrt{t}\left(1 + t\right)\left(1 - t\right)}} \space \mathrm{dt}$$
$$= -\dfrac{1}{4}\large{\int} \normalsize{\dfrac{1}{1 + \left(\sqrt{t}\right)^2}} \space \mathrm{d\left[\sqrt{t}\space\right]} + \dfrac{1}{8}\large{\int} \normalsize{\dfrac{1}{\sqrt{t} + 1}} - \dfrac{1}{\sqrt{t} -1} \space \mathrm{d\left[\sqrt{t}\space\right]}$$
$$\bbox[5px, border: 2px solid black]{=-\dfrac{1}{4}\tan^{-1}{\sqrt{t}} \space + \space \dfrac{1}{8}\ln{\left|\dfrac{1+\sqrt{t}}{\sqrt{t} - 1}\right|} + C} \quad \quad (1)$$
$\newcommand{\answer}[1]{-\dfrac{1}{4}\tan^{-1}{\sqrt{#1}} \space + \space \dfrac{1}{8}\ln{\left|\dfrac{1 + \sqrt{#1}}{\sqrt{#1} - 1}\right|} + C}$


$x = \tan{u} \longleftrightarrow u = \tan^{-1}{x} \\ \therefore \cos{2u} = \dfrac{1 - x^2}{1 + x^2}$

$$\require{AMScd}\begin{CD}(1) = {-\dfrac{1}{4}\tan^{-1}{\sqrt{\cos{2u}}} \space + \space \dfrac{1}{8}\ln{\left|\dfrac{1+\sqrt{\cos{2u}}}{\sqrt{\cos{2u}} - 1}\right|} + C} \\ @VV \large{\cos{2u} \space = \space \frac{1 - x^2}{1 + x^2}} V  \\ \end{CD}$$
$$\bbox[5px, border: 2px solid black]{\answer{\dfrac{1 - x^2}{1 + x^2}}}$$
