Integral $\int \sqrt{x+\sqrt{x^2+2}}dx$ $$\int \sqrt{x+\sqrt{x^2+2}}\ dx$$
I tried various solving methods but I am not coming forward. I unformed the term also to ${x+\sqrt{x^2+2} \over \sqrt{x+\sqrt{x^2+2}}}$  and even multiplied with ${\sqrt{x-\sqrt{x^2+2}} \over \sqrt{x-\sqrt{x^2+2}}}$. 
Trigonometric and hyperbolic substitution didn't help either.
 A: HINT: set $t=\sqrt{x+\sqrt{x^2+2}}$ then we get $x=\frac{t^4-2}{2t^2}$ and we have $$dx=\frac{t^4+2}{t^3}dt$$
A: Let $$\displaystyle I = \int \sqrt{x+\sqrt{x^2+2}}dx\;$$
Now Put $$\displaystyle (x+\sqrt{x^2+2}) = e^{2t}\;$$ Then $$\displaystyle \left(1+\frac{x}{\sqrt{x^2+2}}\right)dx = 2e^{2t}dt$$
So we get $$\displaystyle \left(\frac{e^{2t}}{\sqrt{x^2+2}}\right)dx = 2e^{2t}dt\Rightarrow dx = 2\sqrt{x^2+2}dt$$
Now Using $$\displaystyle \bullet\;  \left(\sqrt{x^2+2}+x\right)\cdot \left(\sqrt{x^2+2}-x\right) = 2$$
So we get $$\displaystyle \left(\sqrt{x^2+2}-x\right) = \frac{2}{e^{2t}}$$
Now $$\displaystyle \sqrt{x^2+2} = \frac{1}{2}\cdot \left(e^{2t}+\frac{2}{e^{2t}}\right)$$
So Integral $$\displaystyle I = \int e^{t}\cdot \left(e^{2t}+\frac{2}{e^{2t}}\right)dt = \int e^{3t}dt+2\int e^{-t}dt$$
So we get $$\displaystyle I = \frac{1}{3}e^{3t}-2e^{-t}+\mathcal{C} = \frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\cdot \left(x+\sqrt{x^2+2}\right)^{-\frac{1}{2}}+\mathcal{C}$$
A: Notice, let $$x+\sqrt{x^2+2}=t^2\implies x=\frac{t^4-2}{2t^2}$$$$ \left(1+\frac{x}{\sqrt{x^2+2}}\right)dx=2tdt\iff \left(\frac{x+\sqrt{x^2+2}}{\sqrt{x^2+2}}\right)dx=2tdt$$$$\implies dx=\frac{t^4+2}{t^3}dt$$
Now, we get
$$\int\sqrt{x+\sqrt{x^2+2}}dx=\int t\frac{t^4+2}{t^3}dt$$
$$=\int \frac{t^4+2}{t^2}dt=\int\left(t^2+\frac{2}{t^2}\right)dt$$
$$=\frac{t^3}{3}-\frac{2}{t}+C$$
$$=\color{}{\frac{(x+\sqrt{x^2+2})\sqrt{x+\sqrt{x^2+2}}}{3}-\frac{2}{\sqrt{x+\sqrt{x^2+2}}}+C}$$
$$=\color{blue}{\frac{(x+\sqrt{x^2+2})^2-6}{3\sqrt{x+\sqrt{x^2+2}}}+C}$$
