Integral $\int \sqrt{x+\sqrt{x^2+2}}dx$

Question

$$\int \sqrt{x+\sqrt{x^2+2}}\ dx$$

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.

Answer 1

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

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

Answer 2

HINT: set $t=\sqrt{x+\sqrt{x^2+2}}$ then we get $x=\frac{t^4-2}{2t^2}$ and we have $$dx=\frac{t^4+2}{t^3}dt$$

Answer 3

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 $$\displaystyle I = \frac{1}{3}e^{3t}-2e^{-t}+\mathcal{C} = \frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\cdot \left(x+\sqrt{x^2+2}\right)^{-\frac{1}{2}}+\mathcal{C}$$