## Calculus 10th Edition

$\int(8x^3+\frac{1}{2x^2})dx=\int(8x^3+\frac{1}{2x^2})dx$
In order to take the derivative of the righthand side of the equation, first rewrite it in order to derive easier: $2x^4-\frac{1}{2x}+C=2x^4-2x^{-1}+C$ Next, take the derivative: $\frac{d}{dx}(2x^4-2x^{-1}+C)=(4)2x^{4-1}-(-1)2x^{-1-1}+0$ $=8x^3+2x^{-2}$ Finally, rewrite the derivative and place it within an integral so that it may be set equal to the lefthand side of the equation: $\int(8x^3+\frac{1}{2x^2})dx=\int(8x^3+\frac{1}{2x^2})dx$