I have to integrate
$$\int \frac{\sqrt{x}}{\sqrt[4]{x^3+1}} \,dx$$
To solve this, I've tried various substitutions including
a) let $x^{3/2}$=$e^{2u},$ which leads to the following integral
$$\frac{4}{3} \int \frac{e^{2x}}{\sqrt [4] {e^{4x}+1}}du$$ which, according to integral calculator.net, has no closed form.
I also tried the following $u$-sub let $u=x^{3/2}.$ However that didn't help much either, so what is the sub, if any that I must use here.
Edit:- My book does give the answer, but not the method, @greg Martin suggested I add that as it may help, so here it is
$\frac{4}{3} (\sqrt[4] {x^3} -\ln(\sqrt[4]{x^3} +1)) +C $
This problem is from N.Piskunov's Differential and integral calculus , Volume 1
Best Answer
The only solution is given in terms of Gauss hypergeometric function. $$\frac 1{\sqrt[4]{t+1}}=\sum_{n=0}^\infty \binom{-\frac{1}{4}}{n} t^n$$ $$\frac{\sqrt{x}}{\sqrt[4]{x^3+1}}=\sum_{n=0}^\infty \binom{-\frac{1}{4}}{n} x^{3 n+\frac{1}{2}}$$ $$\int \frac{\sqrt{x}}{\sqrt[4]{x^3+1}} \,dx=\frac 2 3 x^{3/2}\sum_{n=0}^\infty \frac{\binom{-\frac{1}{4}}{n} }{2 n+1} x^{3 n}=\frac{2}{3} x^{3/2} \,\, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{3}{2};-x^3\right)$$