Question

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

Answer 1

`intu dv=2x^2(lnx^2-1)`

Explanation:

The Integration by Parts formula is shown by;

`intu dv=u v-intv du`

To choose `u` and `dv`, there are several things to be considered;

1. `dx` should be part of `dv`.
2. `dv` should be readily integrated.
3. `u` becomes simpler when differentiated.
4. `intv du` should be simpler than `intu dv`.

When answering this type of question, ensure that `u` choosen can usually differentiates to zero, while `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=lnx^2` and `dv=4x dx`
Then, `du=2/(x)dx` and `v=2x^2`

Using Integration by Parts formula;

`intu dv=u v-intv du`

`intu dv=lnx^2(2x^2)-int(2x^2)(2/x)dx`

`intu dv=lnx^2(2x^2)-int4xdx`

`intu dv=lnx^2(2x^2)-2x^2`

`intu dv=2x^2(lnx^2(1)-1)`

`intu dv=2x^2(lnx^2-1)`