Evaluate
$$\int \frac{dx}{\sqrt{x+\sqrt[3]{x}}}$$
This question was asked in my high school exam, we haven't discussed the solutions yet. I tried substituting $x=\tan^3(\alpha)$ so that I can get $\sqrt {\tan(\alpha) \sec^2(\alpha)}$ in the denominator and then I tried using By-parts but I was still getting an integral which I was not able to solve, kindly don't post the whole solution, just a hint would be sufficient.
Update: Some of my classmates are saying that it should be $\sqrt{x}$ instead of $\sqrt[3]{x}$ in the denominator and solving that is easy, but still I am curious about this integral.
Best Answer
I suspect you've copied the problem incorrectly. Anyway, according to Mathematica:
$$\int \frac{dx}{ \sqrt{x + \sqrt[3]{x}}}= \frac{2 \sqrt{x+\sqrt[3]{x}} \left(\, _2F_1\left(\frac{3}{4},1;\frac{5}{4};-\frac{1}{x^{2/3}}\right)+x^{2/3}\right)}{x^{2/3} }$$
where $F$ is the Hypergeometric function, and
$$\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}} = 2 \sqrt{x}-3 \sqrt[3]{x}+6 \sqrt[6]{x}-6 \log \left(\sqrt[6]{x}+1\right)$$