Answer
$$
f^{\prime}(u)= \frac{u^{2}+\sqrt {u} }{u}=u+u^{-\frac{1}{2}}, \quad f(1)=3
$$
We find that
$$
f(u)= \frac{1}{2}u^{2}+2u^{\frac{1}{2}}+\frac{1}{2}.
$$
Work Step by Step
$$
f^{\prime}(u)= \frac{u^{2}+\sqrt {u} }{u}=u+u^{-\frac{1}{2}}, \quad f(1)=3
$$
The general anti-derivative of $
f^{\prime}(u)= \frac{u^{2}+\sqrt {u} }{u}
$ is
$$
f(u)= \frac{1}{2}u^{2}+2u^{\frac{1}{2}}+C
$$
To determine C we use the fact that $f(1)=3$:
$$
f(1)= \frac{1}{2}(1)^{2}+2(1)^{\frac{1}{2}}+C=3
$$
$ \Rightarrow $
$$
\frac{5}{2}+C=3 \quad \Rightarrow \quad C=\frac{1}{2},
$$
so
$$
f(u)= \frac{1}{2}u^{2}+2u^{\frac{1}{2}}+\frac{1}{2}.
$$
