#blogstopublish

Today, I’m looking at a lovely log integration trick I read about here.

The setup is: you need to integrate something awful like $I=\int 7x^{10}\ln\left(x^3\right) \mathrm dx$.

My initial thought would be (if I didn’t know the trick) to try to do some sort of $u$-substitution and/or integrate by parts. Turns out, that’s not necessary.

Because it’s a straightforward logarithm, it’s easy to manipulate the power. If the argument was $x^{11}$, we’d have an almost immediate function-derivative set-up – and all we’d need to do is multiply it by $\frac{11}{3}$.

Only we can’t just multiply by $\frac{11}{3}$ because that changes the value of the integral. We can multiply by $\frac{3}{11}$ and by $\frac{11}{3}$, though:

$I = \frac{3}{11} \int 7x^{10} \cdot \frac{11}{3}\ln\left(x^{3}\right)\mathrm dx$, which is $I = \frac{3}{11} \int 7x^{10}\ln\left(x^{11}\right)\mathrm dx$ once you fix the power.

It’s already a lot nicer, but we can do better yet: because the derivative of $x^{11}$ is $11x^{10}$, it’d be nice if we could make the 7 into an 11. We do a very similar trick: multiply by $\frac{11}{7}$ to make the constant 11, and also by $\frac{7}{11}$ so we don’t change the value of the integral.

$I = \frac{3}{11} \frac{7}{11} \int \frac{11}{7}\cdot7x^{10}\ln\left(x^{11}\right)\mathrm dx$, which is $I = \frac{21}{121} \int 11x^{10}\ln(x^{11})\mathrm dx$.

Now set $u = x^{11}$ (so $\frac{du}{dx} = 11x^{10}$ and you’ve got $I = \frac{21}{121}\int \frac{ \mathrm du}{ \mathrm dx}\ln(u) \mathrm dx$, or $I= \frac{21}{121}\int \ln(u) \mathrm du$.

You do still need to integrate the logarithm ((My preferred method is by parts, but other options are available))– you get $\frac{21}{121}\left(u\ln(u) - u\right) +C$, or $\frac{21}{121}x^{11}\left(\ln(x^{11})-1\right)+C$.

This, incidentally, is a Very Mathematician Thing: sometimes finding a way to add 0 or multiply by 1 means you can manipulate objects into a better form.