Question

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

Answer 1

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`