# How do you integrate intx^(2/3) *ln x from 1 to 4?

Apr 1, 2015

$\int {x}^{r} \ln x \mathrm{dx}$ is a 'standard' question.

For $r \ne - 1$ integrate by parts.

We don't (most of us) know the integral of $\ln x$, but we do know its derivative, so
Let $u = \ln x$ and $\mathrm{dv} = {x}^{r} \mathrm{dx}$ (in this question $\mathrm{dv} = {x}^{\frac{2}{3}} \mathrm{dx}$

With these choices, we get:

$\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $v = \int {x}^{r} \mathrm{dx} = {x}^{r + 1} / \left(r + 1\right)$ (Here: $\frac{3}{5} {x}^{\frac{5}{3}}$)

$\int u \mathrm{dv} = u v - \int v \mathrm{du} = \frac{3}{5} {x}^{\frac{5}{3}} \ln x - \frac{3}{5} \int {x}^{\frac{5}{3}} \cdot \frac{1}{x} \mathrm{dx}$

$= \frac{3}{5} {x}^{\frac{5}{3}} \ln x - \frac{3}{5} \int {x}^{\frac{2}{3}} \mathrm{dx}$ (You'll always get the same integral fo this kind of problem.)

$= \frac{3}{5} {x}^{\frac{5}{3}} \ln x - \frac{3}{5} \cdot \frac{3}{5} {x}^{\frac{5}{3}} + C$ Or for your definite integral:

$= {\left[\frac{3}{5} {x}^{\frac{5}{3}} \ln x - \frac{9}{25} {x}^{\frac{5}{3}}\right]}_{1}^{4}$

Now do the arithmetic. (remember that $\ln 1 = 0$)