How do you evaluate the integral #int x(lnx)^2dx#?

1 Answer
May 23, 2017

Recursively use integration by parts

Explanation:

#intudv = uv - intvdu#

Let #u = (ln(x))^2 and dv = xdx#, then #du =(2ln(x))/xdx and v = x^2/2#

#intx(ln(x))^2dx = (xln(x))^2/2 - intxln(x)dx#

Integration by parts for the second integral:

Let #u = ln(x) and dv = xdx#, then #du =1/xdx and v = x^2/2#

#intx(ln(x))^2dx = (xln(x))^2/2 - x^2/2ln(x)+1/2intxdx#

#intx(ln(x))^2dx = (xln(x))^2/2 - x^2/2ln(x)+1/4x^2+ C#