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. 

Let your integral be $I$. Then from what you wrote we have 
$$I=x\sqrt{x^2+6}-\int\left(\frac{x^2+6}{\sqrt{x^2+6}}-\frac{6}{\sqrt{x^2+6}}\right)\,dx,$$ which we can rewrite as 
$$I=x\sqrt{x^2+6}-I+6\int \frac{1}{\sqrt{x^2+6}}\,dx.$$ 
Solve for $I$. 
The integral that remains to do is straight Euler substitution $t=x+\sqrt{x^2+6}$.  Then $\frac{dx}{\sqrt{x^2+6}}=\frac{dt}{t}$. 
