Question

Who wants to have a fun problem?

Answer 1

\[\Large \int\limits (27e^{9x}+e^{12x})^{\frac{1}{3}}~~dx \]

Answer 2

Me ME~

Answer 3

Me ME~

Answer 4

would simplifying before integrating make it easier to go?

Answer 5

it becomes e^(3x) * (27+e^(3x))^(1/3) * dx
now sub
u = 27+e^(3x)
du = 3e^(3x)dx

done isnt it ?

Answer 6

You sure about that? @Coolsector

Answer 7

yes

you will have then
u^(1/3)du / 3
so
u^(4/3) / 4

->

(27+e^(3x))^(4/3) / 4

Answer 8

Hm, weird, you got the right answer somehow.
I started with:
\[\Large \int\limits~((e^{9x})(27+e^{3x}))~dx\]

Answer 9

*\[\Large \int\limits\limits~((e^{9x})(27+e^{3x}))^{1/3}~dx \]

Answer 10

well.. that is what i did..
now i took the e^(9x) out of the ( )^(1/3) ..

Answer 11

Wait, nevermind. Sorry about that. I'm half asleep right now and cannot think straight.

Answer 12

oh that is ok.. happens to me a lot :)