What is the antiderivative of the natural logarithm of x?

1 Answer
Feb 24, 2017

Answer:

# 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#