# How do you find the integral of 7x^2ln(x) dx?

Mar 8, 2015

I would use Integretion by Parts: Mar 8, 2015

How: Use integration by parts.
(Any integral of the form $a {x}^{n} \ln x$ can be done by parts.)

Details:
$\int 7 {x}^{2} \ln \left(x\right) \mathrm{dx} = 7 \int {x}^{2} \ln x \mathrm{dx}$

Let $u = \ln x$ and $\mathrm{dv} = {x}^{2} \mathrm{dx}$

This makes $\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $v = \int {x}^{2} \mathrm{dx} = \left(\frac{1}{3}\right) {x}^{3}$ (We'll add the $+ C$ later.)

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$7 \int {x}^{2} \ln x \mathrm{dx} = 7 \left[\ln x \left(\frac{1}{3}\right) {x}^{3} - \int \left(\frac{1}{3}\right) {x}^{3} \cdot \frac{1}{x} \mathrm{dx}\right]$

$= 7 \left[\frac{1}{3} {x}^{3} \ln x - \frac{1}{3} \int {x}^{3} / x \mathrm{dx}\right] = 7 \left[\frac{1}{3} {x}^{3} \ln x - \frac{1}{3} \int {x}^{2} \mathrm{dx}\right]$

$= 7 \left[\frac{1}{3} {x}^{3} \ln x - \frac{1}{3} \left(\frac{1}{3} {x}^{3}\right)\right] + C$

$= \frac{7}{3} {x}^{3} \ln x - \frac{7}{9} {x}^{3} + C$

General
Notice that the solution can be made quite general.
For example: with 5 instead of 3 the problem becomes $\int {x}^{5} \ln x \mathrm{dx}$ which can be solved by the same reasoning to get ; $\frac{1}{6} {x}^{6} \ln x - \frac{1}{36} {x}^{6} + C$

Note: For even better understanding, check the answers by differentiating.