What is the antiderivative of the natural logarithm of x?

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Answer 1 · 2017-02-24T17:27:14

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

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