Good day folks, last couple of days I had been having a good time with indefinite integrals up until stumpled upon these two $$\int \frac{dx}{\sqrt{x+1}+\sqrt[3]{x}}$$ and $$\int \frac{dx}{\sqrt[3]{1+x^2}}$$ after what's been about 2 or 3 days I gave up and decided to use a calculator to help me out. But I tried a bunch of them and they either dont compute or give a result in terms of non-elementary functions existence of which we (as current students) shouldn't even know about, let alone use for solving.
Are they really unsolvable (Well, unsolvable in terms of elementary functions, but those are synonyms at this point) ?
For integral 1, making the substitution $u=x^{1/3},\frac{du}{dx}=\frac1{3u^2}$ as suggested in comments: $$\int_0^y\frac1{\sqrt{x+1}+\sqrt[3]x}\,dx=\int_0^{y^{1/3}}\frac{3u^2}{u+\sqrt{u^3+1}}\,du$$ which is indeed a non-elementary integral (but can be expressed in terms of elliptic integrals – it gets really messy).
Integral 2 is explicitly covered in Byrd and Friedman (274.01), so is also non-elementary and elliptic: $$\int_0^y\frac1{\sqrt[3]{x^2+1}}\,dx=\frac{3(1-\sqrt3)}{2\sqrt[4]3}F(\varphi,m)+3^{5/4}\int_0^{u_1}\frac1{1+\operatorname{cn}u}\,du$$ where $m$ is the the parameter and $$\operatorname{cn}u_1=\cos\varphi=\frac{\sqrt3+1-\sqrt[3]{y^2+1}}{\sqrt3-1+\sqrt[3]{y^2+1}}\qquad m=\frac{2-\sqrt3}4$$ The integral with bounds $[0,u_1]$ above in turn evaluates by B&F 341.53 as $$\int_0^{u_1}\frac1{1+\operatorname{cn}u}\,du=F(\varphi,m)-E(\varphi,m)+\frac{\operatorname{sn}u_1\operatorname{dn}u_1}{1+\operatorname{cn}u_1}$$