Integrating the Natural Logarithm

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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}(xln 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!