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

1 Answer
Jan 11, 2016

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)