By making the change of variable $x^p = y$ we see that $$ \int_0^1 x^{p-1} (1-x)^{q-1} \, dx = {1 \over p} \int_0^1 (1 - y^{1 \over p})^{q-1} \, dy . $$ Hence, we have $$ \int_0^1 (1 - y^{1 \over p})^{q-1} \, dy = p {\Gamma (p) \Gamma (q) \over \Gamma (p + q)} . $$