Question

What is the antiderivative of the natural logarithm of x?

Answer 1

`int \ lnx \ dx =xlnx-x +c`

Explanation:

This answer assumes you know how to integrate by parts.

We start by rewriting `int \ lnx \ dx` as `int \ 1xxlnx \ dx`.

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

We know how to differentiate `lnx`, so we set `u=lnx` and `v'=1`

Integrating `v'` to get `v` gives us `v=x`.
Differentiating `u` to get `u'` give us `u'=1/x`.

We can now substitute this into the formula: `int \ lnx \ dx=xlnx-int \ x1/x \ dx`

This simplifies to `int \ lnx \ dx=xlnx-int \ 1 \ dx`

The integral of `1` is `x`, so `int\ lnx\ dx=xlnx-x + c`