## Thomas' Calculus 13th Edition

$\dfrac{2}{3}x^{3/2}+\dfrac{3}{4}x^{4/3}+C$
As we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Re-write $\int (\sqrt x+\sqrt [3] x) dx$ as $\int (\sqrt x+\sqrt [3] x) dx=\int (x^{1/2}+ x^{1/3}) dx$ Thus, $\int (x^{1/2}+ x^{1/3}) dx=\dfrac{x^{3/2}}{3/2}+\dfrac{x^{4/3}}{4/3}+C=\dfrac{2}{3}x^{3/2}+\dfrac{3}{4}x^{4/3}+C$