Answer
$f(u) = \frac{1}{2}u^2+2\sqrt{u}+\frac{1}{2}$
Work Step by Step
$f'(u) = \frac{u^2+\sqrt{u}}{u} = \frac{u^2}{u}+ \frac{\sqrt{u}}{u} = u+u^{-1/2}$
$f(u) = \int (u+u^{-1/2})~du = \frac{1}{2}u^2+2\sqrt{u}+C$
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:
$f(u) = \frac{1}{2}u^2+2\sqrt{u}+\frac{1}{2}$
