Solving this indefinite integral: $\int{\frac{x\:dx} {(\sqrt{1+x} +\sqrt[3]{1+x} )}} $

Question

I don't know how to solve this integral I tried change of variable but had no results.

$$\int_{0}^{1}{\frac{x\:dx} {(\sqrt{1+x} +\sqrt[3]{1+x} )}} $$

Answer 1

Hint. Just make the change of variable $$ (x+1)=u^6 $$ then $$ \int{\frac{xdx} {(\sqrt{1+x} +(1+x)^{1/3} )}}=6\int{\frac{u^6-1} {u^3+u^2}}u^5\:du $$

Answer 2

Let $1+x=t^6\implies dx=6t^5\ dt$ $$\int \frac{x}{\sqrt {1+x}+\sqrt[3]{1+x}}\ dx$$ $$=\int \frac{(t^6-1)}{t^3+t^2}(6t^5\ dt)$$ $$=6\int \frac{t^3(t^6-1)}{t+1}\ dt$$ $$=6\int \frac{t^3(t^3-1)(t^3+1)}{t+1}\ dt$$ $$=6\int \frac{(t^6-t^3)(t+1)(t^2-t+1)}{t+1}\ dt$$ $$=6\int (t^6-t^3)(t^2-t+1)\ dt$$ $$=6\int (t^8-t^7+t^6-t^5+t^4-t^3)\ dt$$ $$=6\left(\frac{t^9}{9}-\frac{t^8}{8}+\frac{t^7}{7}-\frac{t^6}{6}+\frac{t^5}{5}-\frac{t^4}{4}\right)+C$$ $$=6\left(\frac{(1+x)^{3/2}}{9}-\frac{(1+x)^{4/3}}{8}+\frac{(1+x)^{7/6}}{7}-\frac{(1+x)}{6}+\frac{(1+x)^{5/6}}{5}-\frac{(1+x)^{2/3}}{4}\right)+C$$