How do I find the integral #int(x*ln(x))dx# ?

1 Answer
Aug 18, 2014

We will use integration by parts.

Remember the IBP's formula, which is

#int u dv = uv - int v du#

Let #u = ln x#, and #dv = x dx#. We have chosen these values because we know that the derivative of #ln x# is equal to #1/x#, meaning that instead of integrating something complex (a natural logarithm) we now will end up integrating something pretty easy. (a polynomial)

Thus, #du = 1/x dx#, and #v = x^2 / 2#.

Plugging into the IBP's formula gives us:

#int x ln x dx = (x^2 ln x)/2 - int x^2 / (2x) dx#

An #x# will cancel off from the new integrand:

#int x ln x dx = (x^2 ln x)/2 - int x / 2 dx#

The solution is now easily found using the power rule. Don't forget the constant of integration:

#int x ln x dx = (x^2 ln x)/2 - x^2 / 4 + C#