How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.
Would someone please give me an hint on this ?
Thanks.

Substitute $x=\frac{2-t}{2\sqrt t}$. Then, $dx=-\frac {2+t}{4t\sqrt t}\>dt$ and
\begin{align}
&\int\sqrt{1+x\sqrt{x^2+2}}\>dx \\
= & -\frac18 \int \frac{t+2}{t^2} \sqrt{ 4+4t-t^2}\>dt
= -\frac18 \int \frac{\sqrt{ 4+4t-t^2}}{t-2}d\left( \frac{(t-2)^2}t\right)\\
=& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}- \int \frac1{t\sqrt{ 4+4t-t^2}}dt\\
=& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}+\frac12 \tanh^{-1} \frac{t+2}{\sqrt{ 4+4t-t^2}}+C
\end{align}
