# How do you integrate int lnx by parts?

Nov 28, 2016

$\int \ln x \mathrm{dx} = x \ln x - x + c$

#### Explanation:

This is one of the maths problems where you need to learn a little trick and write $\ln x$ as the product $\left(1\right) \left(\ln x\right)$

If you are studying maths, then you should learn the formula for Integration By Parts (IBP), and practice how to use it:

$\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$, or less formally $\int u \mathrm{dv} = u v - \int v \mathrm{du}$

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

So for the integrand $\left(1\right) \left(\ln x\right)$, hopefully you can see that $\ln x$ simplifies when differentiated and $1$ will become $x$ when integrated, but the logarithm will be eliminated

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

Then plugging into the IBP formula gives us:
$\int \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \mathrm{dx} = \left(u\right) \left(v\right) - \int \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \mathrm{dx}$
$\therefore \int \left(1\right) \left(\ln x\right) \mathrm{dx} = \left(\ln x\right) \left(x\right) - \int \left(x\right) \left(\frac{1}{x}\right) \mathrm{dx}$
$\therefore \int \ln x \mathrm{dx} = x \ln x - \int 1 \mathrm{dx}$
$\therefore \int \ln x \mathrm{dx} = x \ln x - x + c$