Dear Uncle Colin,

Someone has challenged me to solve $\int \sqrt{3x-1+x\sqrt{8x-3}} \dx + \int \sqrt{3x-1-x\sqrt{8x-3}} \dx$. It looks… gnarly. Any ideas?

Getting Nowhere, Asking Researchers Like You

Hi, GNARLY, and thanks for your message! This one caused me a bit of head-scratching before it dropped out really nicely in the end.

The key step is to do the substitution in plain sight: let $u = 8x-3$ in both integrands, so $x = \frac{u+3}{8}$.

The first one becomes $\frac{3}{8}(u+3) - 1 + \frac{1}{8}(u+3)\sqrt{u}$ inside the cube root, or $\frac{1}{8}\left((3u + 9) - 8 + u\sqrt{u} + 3\sqrt{u}\right)$. Simplified down, that’s $\frac{1}{8}\left(1 + 3\sqrt{u} + 3u + u\sqrt{u}\right)$, which is a perfect cube!

Its cube root is $\frac{1}{2}(1+\sqrt{u})$.

The second one goes similarly, only with a minus sign, to give $\frac{1}{2}(1-\sqrt{u})$.

Adding the integrals together gives $\int 1 \dx$ (not $\du$, I haven’t substituted for the differential; I’ve just manipulated algebraically), so the result is $x + c$.

Hope that helps!

- Uncle Colin