Hello, rsanford!
$\displaystyle f'(x) \:= \:\sqrt{x}\,(9 + 5x),\quad f(1) \,=\,11$
Find $\displaystyle f(x)$
We have: .$\displaystyle f'(x) \;=\;9x^{\frac{1}{2}} + 5x^{\frac{3}{2}}$
Integrate: . $\displaystyle f(x)\;=\;\int\left(9x^{\frac{1}{2}} + 5x^{\frac{3}{2}}\right)\,dx \;=\;6x^{\frac{3}{2}} + 2x^{\frac{5}{2}} + C \;=\;2x^{\frac{3}{2}}(3 + x) + C$
Since $\displaystyle f(1) = 11$, we have: . $\displaystyle 2\!\cdot\!1^{\frac{3}{2}}(3 + 1) + C \:=\:11 \quad\Rightarrow\quad C \,=\,3$
Therefore: . $\displaystyle f(x)\;=\;2x^{\frac{3}{2}}(3 + x) + 3$