Integrate $\int\sqrt{x+\sqrt{x^{2}+2}} dx$ . Q)  $\int\sqrt{x+\sqrt{x^{2}+2}} dx$ .
Tried rationalising the numerator twice to get Numerator =-2 but not able to simplify denominator 
The question reduces to (as per my rationalising)
$$\int \frac{-2}{\bigl(x - \sqrt{x^2+2}\bigr)\sqrt{x + \sqrt{x^2+2}}}\,dx$$
 A: Hint:
One could choose :

$$t=\sqrt{x+\sqrt{x^2+2}}$$ then square to get a more meaningful equation$$t^2-x=\sqrt{x^2+2}$$ squaring again and solving for $x$ we get $$x=\frac{t^4-2}{2t^2}$$ 

I'm sure you can take it from here. If not, prompt me for more details.
A: Try the substitution
$$
u=x+\sqrt{x^2+2}.
$$
After simplification, you will probably end up with
$$
\int \frac{1}{2}\sqrt{u}+\frac{1}{u\sqrt{u}}\,du
$$
which you probably can integrate directly(?).
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 $$\displaystyle I = \frac{1}{3}e^{3t}-2e^{-t}+\mathcal{C} = \frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\cdot \left(x+\sqrt{x^2+2}\right)^{-\frac{1}{2}}+\mathcal{C}$$
