# What is the antiderivative of ln(x)^2?

Oct 25, 2016

$\int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} + 2 x + C$

#### Explanation:

You should learn the IBP formula:
$\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

So essentially we are looking for one function that simplifies when it is differentiated, and one that simplifies when integrated (or at least is integrable).

We also need to know that $\int \ln x \mathrm{dx} = x \ln x - x$ (either learn or use IBP):

Let $\left\{\begin{matrix}u = \ln x & \implies & \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \\ \frac{\mathrm{dv}}{\mathrm{dx}} = \ln x & \implies & v = x \ln x - x\end{matrix}\right.$

Then IBP gives;
$\int \ln x \ln x \mathrm{dx} = \ln x \left(x \ln x - x\right) - \int \left(x \ln x - x\right) \frac{1}{x} \mathrm{dx}$
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} - x \ln x - \int \left(\ln x - 1\right) \mathrm{dx}$
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} - x \ln x - \left(x \ln x - x - x\right)$
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} - x \ln x + x \ln x + 2 x$

And so we have:
$\therefore \int {\left(\ln x\right)}^{2} \mathrm{dx} = x {\left(\ln x\right)}^{2} + 2 x + C$