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.