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

1 Answer
Apr 19, 2016

int4x^2ln(x^2)dx=(8x^3(3lnx-1))/9+C

Explanation:

First, rewrite ln(x^2)=2lnx using the rule that ln(a^b)=b*lna.

This gives us the integral int4x^2(2lnx)dx=8intx^2lnxdx.

For this, recalling that integration by parts takes the form intudv=uv-intvdu, we let

u=lnx" "=>" "(du)/dx=1/xdx" "=>" "du=1/xdx

dv=x^2dx" "=>" "intdv=intx^2dx" "=>" "v=x^3/3

This gives us:

8intx^2lnx=8[(x^3/3)lnx-intx^3/3(1/x)dx]

=(8x^3lnx)/3-8intx^2/3dx

=(8x^3lnx)/3-(8x^3)/9+C

Which can also be written as

int4x^2ln(x^2)dx=(8x^3(3lnx-1))/9+C