<s>[INST] How to calculate $\int \sqrt{x^2+6}\,dx$? How to calculate $\int \sqrt{x^2+6}\,dx$,  by using Euler substitution and with to use of the formula : $\int u\,dv = vu - \int v\,du $. 
note: what I mean by Euler substitution: is when we have a Integrand like $ \sqrt{x^2+1}$, then we can use the trick of substituting $t= x + \sqrt{x^2+1}$. 
and that gives us: $dt=\frac{t}{\sqrt{x^2+1}} \, dx$, which can be helpful while solving the questions.
here I supposed that $u=\sqrt{x^2+6}$, and $du= \frac{x}{\sqrt{x^2+6}}\,dx$
then: $\int u\,dv = x\sqrt{x^2+6} - \int \frac{x^2}{\sqrt{x^2+6}}\,dx $, and then I got stuck with the latter integral. trying to substitute $t= x + \sqrt{x^2+6}$ in this case didn't really help. how can I do it $especially$ in this way? I know there are might be a lot of creative solutions, but I want some help continuing that direction. 
 [/INST] For the integral 
\begin{align}
I = \int \sqrt{ x^{2} + 6 } \, dx
\end{align}
let $x = \sqrt{6} \sinh(t)$ to obtain $dx = \sqrt{6} \cosh(t) dt$ and 
\begin{align}
I &= \int \sqrt{6} \, \sqrt{ 6 ( 1 + \sinh^{2}(t))} \, \cosh(t) \, dt \\
&= 6 \int \cosh^{2}(t) \, dt \\
&= 3 \int ( 1 + \cosh(2t)) \, dt \\
&= 3 \left[ t + \frac{\sinh(2t)}{2} \right] \\
&= 3 \left[ \sinh^{-1}\left( \frac{x}{\sqrt{6}} \right) + \frac{ x \, \sqrt{x^{2} + 6} }{ 6} \right]
\end{align}
Notes: Reading the question properly this solution would have been different. With that being said the presentation here is an alternate method of producing a solution to the proposed integral.
</s>