# How do you integrate int 4 x^2 ln x^2 dx  using integration by parts?

Apr 19, 2016

$\int 4 {x}^{2} \ln \left({x}^{2}\right) \mathrm{dx} = \frac{8 {x}^{3} \left(3 \ln x - 1\right)}{9} + C$

#### Explanation:

First, rewrite $\ln \left({x}^{2}\right) = 2 \ln x$ using the rule that $\ln \left({a}^{b}\right) = b \cdot \ln a$.

This gives us the integral $\int 4 {x}^{2} \left(2 \ln x\right) \mathrm{dx} = 8 \int {x}^{2} \ln x \mathrm{dx}$.

For this, recalling that integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$, we let

$u = \ln x \text{ "=>" "(du)/dx=1/xdx" "=>" } \mathrm{du} = \frac{1}{x} \mathrm{dx}$

$\mathrm{dv} = {x}^{2} \mathrm{dx} \text{ "=>" "intdv=intx^2dx" "=>" } v = {x}^{3} / 3$

This gives us:

$8 \int {x}^{2} \ln x = 8 \left[\left({x}^{3} / 3\right) \ln x - \int {x}^{3} / 3 \left(\frac{1}{x}\right) \mathrm{dx}\right]$

$= \frac{8 {x}^{3} \ln x}{3} - 8 \int {x}^{2} / 3 \mathrm{dx}$

$= \frac{8 {x}^{3} \ln x}{3} - \frac{8 {x}^{3}}{9} + C$

Which can also be written as

$\int 4 {x}^{2} \ln \left({x}^{2}\right) \mathrm{dx} = \frac{8 {x}^{3} \left(3 \ln x - 1\right)}{9} + C$