Question

Could anyone help me with this integration problem please?

Answer 1

\[\int\limits_{?}dx/(2-\sqrt[3]x)\]

Answer 2

I can't find a substitution that fits to an inverse trig :/

Answer 3

At first glance I don't know how to approach this particular problem. I've written it down. My suggestion is that you move on, if at all possible, and come back to this problem later. Is that feasible for you?

Answer 4

Please note that \[2=\sqrt[3]{8}\]

Answer 5

That's absolutely fine, thanks :) it's actually a definite integral between 1 and 0 and the answer is 12ln2 - 15/2 !

Answer 6

and so your denominator becomes \[\sqrt[3]{8}-\sqrt[3]{x}\]

Answer 7

Don't know yet whether that will help, but thought I'd share this observation. Please move on to another problem; you (or I) could bump this problem to the top later.

Answer 8

Thanks a lot, i'll do that :)

Answer 9

hi.

Answer 10

Nice name :D

Answer 11

Hehe Thanks I was shocked at first when you and I had the same username. I was like whaaat. But It's okay. haha.

Answer 12

Glad two are friends, or may become friends. But please enjoy your friendship through private messages, not here in public, OK? See that little envelope in the upper, left-hand corner of the screen? Click on that to send messages that won't be seen by all other users.

Answer 13

Try \[u=2-\sqrt[3]{x}\]\[du=-\frac{1}{3\sqrt[3]{x^2}}dx\]\[-3x^{2/3}du=dx\]\[-3(2-u)^2du=dx\]\[\int\limits _{0}^{10}\frac{-3(2-u)^2}{u}du\]

Sometimes you just gotta be brave and try some stuff out! If it doesn't work, just keep trying random things, there are only so many options and if your teacher isn't a jerk it won't be too crazy! =)

Answer 14

Thanks so much! Massive help :)

Answer 15

Glad I could help!