<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] \begin{align}
t & = x + \sqrt{x^2+6} \\[8pt]
dt & = \left( 1 + \frac x{\sqrt{x^2+6}} \right)\,dx = \frac t {\sqrt{x^2+6}} \,dx \\[8pt]
\frac{dt} t & = \frac{dx}{\sqrt{x^2+6}}
\end{align}
\begin{align}
t & = x+\sqrt{x^2+6} \\
2x-t & = x - \sqrt{x^2+6} \\[6pt]
\text{Hence, }t(2x-t) & = -6 \\
0 & = t^2 - 2xt -6 \\[6pt]
x & = \frac{t^2-6}{2t} \\[6pt]
\sqrt{x^2+6} & = \frac{t^2+6}{2t} \\[6pt]
dx & = \sqrt{x^2+6}\,\frac{dt} t = \frac{(t^2+6)\,dt}{2t^2}
\end{align}
Therefore
$$
\int \sqrt{x^2+6}\  dx = \int \frac{t^2+6}{2t} \cdot \frac{(t^2+6)\,dt}{2t^2} \text{ etc.}
$$
</s>