<s>[INST] Shortcuts for integrating $\int \sqrt{x^2-36} \,dx$ Is there any simple method to solve this integral

$$\int \sqrt{x^2-36} \,dx$$

Because my method is a bit complicated and long: 
 [/INST] I think integration by parts is the shortest way here. Let $u = \sqrt{x^2 - 36}$ and $\mathrm dv = \mathrm dx$. Then,
$$\begin{align}
\int\sqrt{x^2 - 36}\,\mathrm dx &= x\sqrt{x^2 - 36} - \int\frac{x^2}{\sqrt{x^2 - 36}}\,\mathrm dx =\\[0.3em]
&= x\sqrt{x^2 - 36} - \int\frac{x^2 - 36 + 36}{\sqrt{x^2 - 36}}\,\mathrm dx =\\[0.3em]
&= x\sqrt{x^2 - 36} - \color{green}{\int\sqrt{x^2 - 36}\,\mathrm dx} - 36\int\frac1{\sqrt{x^2 - 36}}\,\mathrm dx=\\[0.3em]
&= \frac x2\sqrt{x^2 - 36} - 18\int\frac1{\sqrt{x^2 - 36}}\,\mathrm dx =\tag{$\star$}\\[0.5em]
&= \frac x2\sqrt{x^2 - 36} - 18\operatorname{arccosh}\frac x6 + C'
\end{align}$$
In step $(\star)$ we moved the highlighted term to the LHS and divided by $2$. Then we used the fact that
$$\frac{\mathrm d}{\mathrm dx}\operatorname{arccosh} x = \frac1{\sqrt{x^2 - 1}}.$$
You can also rewrite the result in another form, using $\operatorname{arccosh} x$ definition:
$$\operatorname{arccosh} x = \ln\left(x + \sqrt{x^2 - 1}\right);$$
in this case, we can write
$$\begin{align}
\int\sqrt{x^2 - 36}\,\mathrm dx &= \frac x2\sqrt{x^2 - 36} - 18\ln\left(\frac x6 + \frac16\sqrt{x^2 - 36}\right) + C' =\\[0.3em]
&= \boxed{\displaystyle\frac x2\sqrt{x^2 - 36} - 18\ln\left(x + \sqrt{x^2 - 36}\right) + C}
\end{align}$$
</s>