Calculating $\int \sqrt{1 + x^{-2}}dx$ I would like to find
$$\int \sqrt{1 + x^{-2}}dx$$
I have found that it is equivalent to
$$
\int \frac{\sqrt{1 + x^2}}{x}dx
$$
but I am not sure what to do about it. With trig substitution $x = \tan(x)$ I get
$$
\int \frac{1}{\sin(\theta)\cos^2(\theta)}d\theta
$$
but that seems to be a dead end.
 A: $$
\int \frac{1}{\sin(\theta)\cos^2(\theta)}d\theta
$$
$$
=\int \frac{\sin(\theta)}{\cos^2(\theta)(1-\cos^2(\theta))}d\theta
$$
then use $u=\cos(\theta)$ and partial fractions on the result.
A: Let $x^2+1=u^2$, so that $2x\,dx=2u\,du$, which implies 
$${dx\over x}={u\,du\over x^2}={u\over u^2-1}du$$  
Then
$$\int{\sqrt{1+x^2}\over x}dx=\int{u^2\over u^2-1}du={1\over2}\int\left(2+{1\over u-1}-{1\over u+1} \right)du=u+{1\over2}\ln\left(u-1\over u+1 \right)+C\\=\sqrt{x^2+1}+{1\over2}\ln\left(\sqrt{x^2+1}-1\over\sqrt{x^2+1}+1 \right)+C$$
A: and
$$
\int \frac{1}{\sin(\theta)\cos^2(\theta)}d\theta=
\int \frac{\sin\theta}{\sin^2(\theta)\cos^2(\theta)}d\theta=
\int \frac{-d(\cos\theta)}{(1-\cos^2\theta)\cos^2(\theta)}d\theta=...
$$
A: $$\int \sqrt{1 + x^{-2}}dx=\int \frac{\sqrt{1 + x^2}}{x}dx$$
With substitution $u=x^{2}$, $du=2xdx$.
$$\int \frac{\sqrt{1 + x^2}}{x}dx=\int \frac{\sqrt{u+1}}{2u}du=\frac{1}{2} \int \frac{\sqrt{u+1}}{u}du $$
With substitution $v=\sqrt{u+1}$, $dv=\frac{1}{2v}du$.
\begin{align}
\frac{1}{2} \int \frac{u+1}{u}du
&=\frac{1}{2} \int \frac{2v^{2}}{v^{2}-1}dv\\[6px]
&=v-\operatorname{arctanh} (v)+C\\[6px]
&=\sqrt{x^{2}+1}-\operatorname{arctanh}(\sqrt{x^{2}+1})+C.
\end{align}
A: I get $\int \frac{sec \theta}{tan \theta}d \theta$ from your simplification
Which gives $\int \frac{1/cos \theta}{sin \theta / cos \theta}d \theta$
= $\int \frac{1}{sin \theta}d \theta$
= $\int csc \theta d \theta$
which integrates similarly to $sec\  \theta$
and then the fun of back-substituting.
A: Set $x^{-1}=\sinh t$, so $\sqrt{1+x^{-2}}=\cosh t$. Then
$$
dx=-\frac{\cosh t}{\sinh^2t}\,dt
$$
and the integral becomes
$$
-\int\frac{\cosh^2t}{\sinh^2t}\,dt=
-\int\frac{1+\sinh^2t}{\sinh^2t}\,dt=\frac{\cosh t}{\sinh t}-t+c=
\sqrt{1+x^2}-\operatorname{arsinh}\frac{1}{x}+c
$$
You can find a more explicit expression for $\operatorname{arsinh}\frac{1}{x}$ by setting
$$
\frac{1}{x}=\frac{e^t-e^{-t}}{2}
$$
or
$$
xe^{2t}-2e^t-x=0
$$
so
$$
e^t=\frac{1+\sqrt{1+x^2}}{x}
$$
The final antiderivative is
$$
\sqrt{1+x^2}-\log\frac{1+\sqrt{1+x^2}}{x}+c
$$
