Integrating the Natural Logarithm « Calculus: Problems and Solutions

Integrating the Natural Logarithm

July 20, 2009 · 2 Comments

You’ll probably see this integral someday:

$\displaystyle \int \ln x\,dx$ .

It looks so simple, and you think, “gee, I probably was supposed to memorize that” or “oh I can do that, it looks so easy.” But then you don’t remember the antiderivative and get stuck. Cause, heck, what can you do with just $\ln x$ anyway? It doesn’t decompose into anything nicer.

The trick is to use integration by parts. Let’s look at it backwards:

If we were really smart or really lucky, we might approach this problem by saying

$\displaystyle \frac{d}{dx}(x\ln x) = \ln x + 1$ .

(Is this really true? Use the product rule to show it.) Then we can integrate both sides:

$\displaystyle \int\frac{d}{dx}(x\ln x)\,dx = \int\ln x\,dx + \int 1\,dx$ .

Now using the Fundamental theorem of calculus, the integral on the left is just equal to $x\ln x$ . Solving for $\int\ln x\,dx$ , we get

$\displaystyle \int\ln x = x\ln x - x + C$ .

Yay!

Categories: calculus
Tagged: integration, integration by parts, ln

2 responses so far ↓

Integration by Parts « Calculus: Problems and Solutions // August 5, 2009 at 9:30 am | Reply

[...] 5, 2009 · Leave a Comment When we looked at integrating the natural logarithm, we used a little trick: we took the product [...]
David Woodford // November 19, 2009 at 5:53 am | Reply

Interesting, I was taught to do it by writing lnx = 1lnx but this way seems much simpler.
Great site in general,
Dave

Calculus: Problems and Solutions