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

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Answer 1 · 2016-04-19T21:58:12

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

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#