# How do you integrate int lnx dx  using integration by parts?

Jan 5, 2016

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

#### Explanation:

The formula for integration by parts, just for reference:

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

It's not all that obvious what we should set $u$ and $\mathrm{dv}$ as, since we only have one function here.

However, let's set $u = \ln x$. Then $\mathrm{du} = \frac{1}{x} \mathrm{dx}$.

The real trick is to set $\mathrm{dv} = \mathrm{dx}$, making $v = x$. This will allow us to get rid of the $\ln$ and simply be left with a polynomial to integrate.

Plugging into our formula we have

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

$\frac{x}{x}$ is just one:

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

And we know $\int \mathrm{dx}$ turns out to be $x$. Don't forget the constant of integration:

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