Integral of e^(x^(1/3))? I'd imagine some combination of integration by parts and u-substitution are required, but it's not apparent at the moment. An explanation would be ten times more valuable than just an answer.

Question

zepdrix · Answer

Hmm let's try a u-sub and see if it will work...

anonymous · Answer

with what, though? The obvious thing to try would be u=cube root of x, but we don't have a du in there.

zepdrix · Answer

$$\Large\bf\sf \int e^{x^{1/3}}\;dx$$
$$\Large\bf\sf u=x^{1/3}$$$$\Large\bf\sf du=\frac{1}{3}x^{-2/3}\;dx\qquad\to\qquad 3x^{2/3}du=dx$$Which we can write as,$$\Large\bf\sf 3(x^{1/3})^2du=dx$$$$\Large\bf\sf 3u^2\;du =dx$$

zepdrix · Answer

We have to uhhh, "make" a du.
It's a little tricky.

anonymous · Answer

OK! That makes sense!

zepdrix · Answer

Yah I'll bet parts works nicely from there :)

anonymous · Answer

so now we have $$3u^2e^udu$$

anonymous · Answer

which works pretty well with integration by parts, I think.

anonymous · Answer

$$\int\limits3u^2e^udu = 3u^2e^u-6ue^u+6e^u$$

anonymous · Answer

Thanks! I'll remember to try that trick next time I'm stuck - is there a formal name for it?

zepdrix · Answer

Hmm I'm not sure!
It's just a neat little tricky where you sometimes have to use your `u` within your `du`.