<s>[INST] 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.
 [/INST] 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 $$\displaystyle I = \frac{1}{3}e^{3t}-2e^{-t}+\mathcal{C} = \frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\cdot \left(x+\sqrt{x^2+2}\right)^{-\frac{1}{2}}+\mathcal{C}$$
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