## Calculus: Early Transcendentals 8th Edition

$f(u) = \frac{1}{2}u^2+2\sqrt{u}+\frac{1}{2}$
$f'(u) = \frac{u^2+\sqrt{u}}{u} = \frac{u^2}{u}+ \frac{\sqrt{u}}{u} = u+u^{-1/2}$ $f(u) = \int (u+u^{-1/2})~du = \frac{1}{2}u^2+2\sqrt{u}+C$ Note that $f(1) = 3$ We can find the value of $C$: $f(1) = \frac{1}{2}(1)^2+2\sqrt{1}+C = 3$ $\frac{1}{2}+2+C = 3$ $C = \frac{1}{2}$ Therefore: $f(u) = \frac{1}{2}u^2+2\sqrt{u}+\frac{1}{2}$