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

Question

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)

$$\int \frac{-2}{\bigl(x - \sqrt{x^2+2}\bigr)\sqrt{x + \sqrt{x^2+2}}}\,dx$$

Answer 1

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

Answer 2

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(?).

Answer 3

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.