## Thomas' Calculus 13th Edition

$4y^2-\dfrac{2y^{3/4}}{3/4} +C=4y^2-\dfrac{8y^{3/4}}{3} +C$
As we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Simpify. $\int (8y-\dfrac{2}{y^{1/4}})dy$ Thus, $\int (8y-\dfrac{2}{y^{1/4}})dy=4y^2-\dfrac{2y^{3/4}}{3/4} +C=4y^2-\dfrac{8y^{3/4}}{3} +C$