Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 40

Evaluate $\displaystyle \int x^{\frac{3}{2}}$. Illustrate and check whether your answer is reasonable by graphing both the function and its antiderivative suppose that $c = 0$.

By using integration by parts, if we let $u = \ln x$ and $dv = x^{\frac{3}{2}} dx$, then
$\displaystyle du = \frac{1}{x} dx \qquad v = \frac{2}{5} x^{\frac{5}{2}}$

So,

$
\begin{equation}
\begin{aligned}
\int x^{\frac{3}{2}} \ln x dx = uv - \in v du &= \frac{2}{5} x^{\frac{5}{2}} \ln x - \int \frac{2}{5} x^{\frac{5}{2}} \left( \frac{1}{x} \right)\\
\\
&= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \int x^{\frac{5}{2}-1} dx\\
\\
&= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \int x^{\frac{3}{2}} dx\\
\\
&= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \left[ \frac{x^{\frac{5}{2}}}{\frac{5}{2}} \right] + c\\
\\
&= \frac{2}{5} x^{\frac{5}{2}} \left[ \ln x - \frac{2}{5} \right] + c
\end{aligned}
\end{equation}
$





We can see from the graph that our answer is reasonable, because the graph of the anti-derivative $f$ is increasing when $f'$ is positive. On the other hand, the graph of $f$ is decreasing when $f'$ is negative.