# What is int x^2lnxdx?

Nov 9, 2015

Use $f \left(x\right) = \ln \left(x\right)$, $g ' \left(x\right) = {x}^{2}$.
The result is: $\frac{1}{3} {x}^{3} \ln \left(x\right) - \frac{1}{9} {x}^{3}$.

#### Explanation:

Use "integration by parts".

The rule is:
$\int f \left(x\right) g ' \left(x\right) \mathrm{dx} = f \left(x\right) g \left(x\right) - \int f ' \left(x\right) g \left(x\right) \mathrm{dx}$

A very important decision is now which part of your product is $f \left(x\right)$ and which one is $g ' \left(x\right)$ since later, you will need to differentiate one of the terms and integrate the other one. With a wrong choice, you might never find a solution.

Here, the choice is not very hard: you really want to differentiate and not integrate $\ln \left(x\right)$ (trust me on that ;-) ), so choose $f \left(x\right) = \ln \left(x\right)$ and $g ' \left(x\right) = {x}^{2}$.

Now, you need to compute $f ' \left(x\right) = \frac{1}{x}$ and $g \left(x\right) = \frac{1}{3} {x}^{3}$.

At this point you are ready to apply the rule.

$\int {x}^{2} \ln \left(x\right) \mathrm{dx} = \ln \left(x\right) \cdot \frac{1}{3} {x}^{3} - \int \frac{1}{x} \cdot \frac{1}{3} {x}^{3} \mathrm{dx}$
$= \frac{1}{3} {x}^{3} \ln \left(x\right) - \frac{1}{3} \int {x}^{2} \mathrm{dx}$
$= \frac{1}{3} {x}^{3} \ln \left(x\right) - \frac{1}{3} \cdot \left(\frac{1}{3} {x}^{3}\right)$
$= \frac{1}{3} {x}^{3} \ln \left(x\right) - \frac{1}{9} {x}^{3}$
or, if you wish,
$= \frac{1}{3} {x}^{3} \left(\ln \left(x\right) - \frac{1}{3}\right)$