How do you integrate #int 4 x^2 ln x^2 dx # using integration by parts?
1 Answer
Apr 19, 2016
Explanation:
First, rewrite
This gives us the integral
For this, recalling that integration by parts takes the form
#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#