An analytical solution of a tricky integral

Question

Can anyone propose a method (some methods) to determine the following indefinite integral?

$$I=\int\frac{\textrm{d}x}{\sqrt[3]{\sqrt{1+x^{2}}-x}}.$$

I think an analytical solution should be possible...

Answer 1

Hint: Substitute $$u=\sqrt{x^2+1}-x$$ then we have $$du=\left(\frac{x}{\sqrt{x^2+1}}-1\right)dx$$ then you will get $$\int \frac{-u^2-1}{2u^{7/3}}du$$ Can you finish? This is $$\int \left(\frac{-u^2}{2u^{7/3}}-\frac{1}{2u^{7/3}}\right)du$$

Answer 2

Another possibility should be substituting $x = \sinh(y)$ and using $1+ \sinh(y)^2 = \cosh(y)^2$ and $\cosh(y) - \sinh(y) = e^{-y}$.

Answer 3

Using the substitution $x = \sinh t$, we have $dx = \cosh t dt$.

\begin{align} I &= \int\frac{dx}{\sqrt[3]{\sqrt{x^2 + 1}-x}}\\ &= \int\frac{\cosh t}{\sqrt[3]{\cosh t - \sinh t}}\,dt\\ &= \int e^{t/3}\cosh t \,dt\\ &= \frac{1}{2}\int\left(e^{4t/3} + e^{-2t/3}\right)\,dt\\ &= \frac{1}{2}\left(\frac{3}{4}e^{4t/3} - \frac{3}{2}e^{-2t/3}\right) + C \end{align}

Recalling our substitution, which implies that $x = \operatorname{arsinh}t$, and using that

$$\exp(\beta\operatorname{arsinh}x) = \left(\sqrt{x^2+1}+x\right)^\beta$$

we obtain that

$$I = \frac{3}{8}\left(\sqrt{x^2+1}+x\right)^{4/3} - \frac{3}{4}\left(\sqrt{x^2+1}+x\right)^{-2/3} + C$$