Answer
$\displaystyle \int f(x)=\frac{2}{25}x^{5/2}(5\ln x-2)+C$
Reasonable, because $f(x)$ behaves as $F'(x)$ should.
Work Step by Step
$f(x)=x^{3/2}\ln x$
$I=\displaystyle \int f(x)dx$ = by parts,
$\displaystyle \left[\begin{array}{ll}
u=\ln x & dv=x^{3/2}dx\\
& \\
du=\frac{dx}{x} & v=\frac{2}{5}x^{5/2}
\end{array}\right],\quad\int udv=uv-\int vdu$
$I==\displaystyle \frac{2}{5}x^{5/2}\ln x-\int\frac{2}{5}x^{5/2}\cdot\frac{dx}{x}$
$=\displaystyle \frac{2}{5}x^{5/2}\ln x-\frac{2}{5}\int x^{3/2}dx$
$=\displaystyle \frac{2}{5}x^{5/2}\ln x-\frac{2}{5}(\frac{2}{5}x^{5/2})+C$
$=\displaystyle \frac{2}{5}x^{5/2}\ln x-\frac{4}{25}x^{5/2}+C$
$=\displaystyle \frac{2}{25}x^{5/2}(5\ln x-2)+C$
Graphing (F is blue, f is red, C=0)
we recognize the characteristics of $f(x)=F'(x)$:
As $F$ decreases (from the left towards $x=1,\ f$ (the derivative) is negative.
When F reaches its minimum, f is zero, changing from negative to positive.
As F rises, f is positive.
So, the answer is reasonable.
