How do you find the integral of #f(x)=(ln6x)^2# using integration by parts?

1 Answer
Nov 29, 2016

The general rule of integration by parts is:

#int u*dv = u*v-int v*du#

Take #u(x) = (ln 6x)^2# and #v(x)=x#

#int (ln6x)^2dx=x(ln 6x)^2 - int x *d ((ln 6x)^2)=x(ln 6x)^2 - int x((2ln 6x)*1/(6x)*6) dx=x(ln 6x)^2-int 2ln6x dx#

Iterating:

#int ln(6x) dx = x ln(6x) -int x d(ln(6x)) = xln(6x) -int x(1/(6x)*6)dx = int dx = xln(6x) - x#

Putting it together:

#int (ln6x)^2dx = x(ln 6x)^2 - 2xln(6x) +2x#