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

Aug 18, 2014

We will use integration by parts.

Remember the IBP's formula, which is

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

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

Thus, $\mathrm{du} = \frac{1}{x} \mathrm{dx}$, and $v = {x}^{2} / 2$.

Plugging into the IBP's formula gives us:

$\int x \ln x \mathrm{dx} = \frac{{x}^{2} \ln x}{2} - \int {x}^{2} / \left(2 x\right) \mathrm{dx}$

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

$\int x \ln x \mathrm{dx} = \frac{{x}^{2} \ln x}{2} - \int \frac{x}{2} \mathrm{dx}$

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

$\int x \ln x \mathrm{dx} = \frac{{x}^{2} \ln x}{2} - {x}^{2} / 4 + C$