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.

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 $$\int\sqrt{x+\sqrt{x^2+2}}dx=\int t\frac{t^4+2}{t^3}dt$$ $$=\int \frac{t^4+2}{t^2}dt=\int\left(t^2+\frac{2}{t^2}\right)dt$$ $$=\frac{t^3}{3}-\frac{2}{t}+C$$ $$=\color{}{\frac{(x+\sqrt{x^2+2})\sqrt{x+\sqrt{x^2+2}}}{3}-\frac{2}{\sqrt{x+\sqrt{x^2+2}}}+C}$$

$$=\color{blue}{\frac{(x+\sqrt{x^2+2})^2-6}{3\sqrt{x+\sqrt{x^2+2}}}+C}$$