## Calculus 10th Edition

$\int (\sqrt[4]{x^3}+1)dx=\dfrac{4}{7\sqrt[4]{x}}+x+C.$
$\int (\sqrt[4]{x^3}+1)dx=\int x^{\frac{3}{4}}dx+\int1 dx=\dfrac{4}{7\sqrt[4]{x}}+x+C.$ By differentiating, we get the same integrand we started with.