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. 
 A: 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 $\frac{dx}{\sqrt{x^2+6}}=\frac{dt}{t}$. 
A: \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
$$
\int \sqrt{x^2+6}\  dx = \int \frac{t^2+6}{2t} \cdot \frac{(t^2+6)\,dt}{2t^2} \text{ etc.}
$$
A: 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.
A: Forget for a second about stupid number 6. You want a substitution which will make $x^2+1$ a perfect square. Let see. Euler seems overkill for something like that. Recall  
\begin{equation}
    \cosh^2(x)-\sinh^2(x)=1
\end{equation}
where $\cosh(x)$ and $\sinh(x)$ stand for hyperbolic function. It should be pretty obvious that you need something like
\begin{equation}
   t=\sqrt{6}\sinh(x)
\end{equation} 
