TITLE: Calculating $\int \sqrt{1 + x^{-2}}dx$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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
$$
\sqrt{1+x^2}-\log\frac{1+\sqrt{1+x^2}}{x}+c
$$