# What is the antiderivative of the natural logarithm of x?

Feb 24, 2017

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

#### Explanation:

This answer assumes you know how to integrate by parts.

We start by rewriting $\int \setminus \ln x \setminus \mathrm{dx}$ as $\int \setminus 1 \times \ln x \setminus \mathrm{dx}$.

This is now a product so we can integrate it by parts using the formula: $\int \setminus v ' u = u v - \int \setminus u ' v$

We know how to differentiate $\ln x$, so we set $u = \ln x$ and $v ' = 1$

Integrating $v '$ to get $v$ gives us $v = x$.
Differentiating $u$ to get $u '$ give us $u ' = \frac{1}{x}$.

We can now substitute this into the formula: $\int \setminus \ln x \setminus \mathrm{dx} = x \ln x - \int \setminus x \frac{1}{x} \setminus \mathrm{dx}$

This simplifies to $\int \setminus \ln x \setminus \mathrm{dx} = x \ln x - \int \setminus 1 \setminus \mathrm{dx}$

The integral of $1$ is $x$, so $\int \setminus \ln x \setminus \mathrm{dx} = x \ln x - x + c$