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

Jan 11, 2016

$\int u \mathrm{dv} = 2 {x}^{2} \left(\ln {x}^{2} - 1\right)$

#### Explanation:

The Integration by Parts formula is shown by;

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

To choose $u$ and $\mathrm{dv}$, there are several things to be considered;

1. $\mathrm{dx}$ should be part of $\mathrm{dv}$.
2. $\mathrm{dv}$ should be readily integrated.
3. $u$ becomes simpler when differentiated.
4. $\int v \mathrm{du}$ should be simpler than $\int u \mathrm{dv}$.

When answering this type of question, ensure that $u$ choosen can usually differentiates to zero, while $\mathrm{dv}$ is easy to integrate.

Also, choose $u$ in this order;
LIPET : L ogs, I nverse trig., P olynomial, E xponential, T rig.

In this question;
Let $u = \ln {x}^{2}$ and $\mathrm{dv} = 4 x \mathrm{dx}$
Then, $\mathrm{du} = \frac{2}{x} \mathrm{dx}$ and $v = 2 {x}^{2}$

Using Integration by Parts formula;

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

$\int u \mathrm{dv} = \ln {x}^{2} \left(2 {x}^{2}\right) - \int \left(2 {x}^{2}\right) \left(\frac{2}{x}\right) \mathrm{dx}$

$\int u \mathrm{dv} = \ln {x}^{2} \left(2 {x}^{2}\right) - \int 4 x \mathrm{dx}$

$\int u \mathrm{dv} = \ln {x}^{2} \left(2 {x}^{2}\right) - 2 {x}^{2}$

$\int u \mathrm{dv} = 2 {x}^{2} \left(\ln {x}^{2} \left(1\right) - 1\right)$

$\int u \mathrm{dv} = 2 {x}^{2} \left(\ln {x}^{2} - 1\right)$