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

Question

2948 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-11T13:38:53

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

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;

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)#