How to find the integral $\int_0^1 \frac{x^{1/2}}{1+x^{1/3}}dx$? How to find $$\int_0^1 \frac{x^{1/2}}{1+x^{1/3}}dx$$ ?
My attempt: I made $t^6=x$ and got $\displaystyle \int_0^1 \frac{6t^8}{1+t^2}dt$ and got stuck.

Hint write it as $6t^8=6t^8+6t^6-6t^6$ then by dividing it by denominator you get it as $6-\frac {6t^6}{1+t^2} $
 Now write the next part of $6t^6=6t^6-6t^4+6t^4$...and continue till $t^2$ to solve an easy integrable function.
