Question

Hey ! what is the integral of x^(1/3)/x^(1/3)-1

Answer 1

substitute x^1/3 = u

Answer 2

\[\int\limits_{}^{}\frac u {u-1}\]

Answer 3

\[\int\limits_{?}^{?} \sqrt[3]{x}\div \sqrt[3]{x}-1\]

Answer 4

yeah

Answer 5

how would u find the deriv of u over u = or - n

Answer 6

use the quotient rule or something. I don't remember, I did this in high school

Answer 7

let me find it and paste it here

Answer 8

alrighty..cool:)

Answer 9

u/(u-1) =1 + 1/(u-1)

Answer 10

ohh...how did u get tht?

Answer 11

there is a technique for it but I dont know that, I just always look at it and think :-)

Answer 12

\[\int\limits \frac{x^{1/3}}{x^{1/3}-1}dx\]
Do a u-sub:
u=x^(1/3)=>x=u^3
dx=3u^2du
\[\int\limits \frac{(u)(3u^2)du}{u-1}=3 \int\limits \frac{u^3}{u-1}du\]
From here, you can do long division. Then it should fall out.

Answer 13

Let me know if you want me to take it further.

Answer 14

take it further plz:)

Answer 15

u=x^(1/3)
du=1/3 x^(-2/3) dx
isnt this the next step?

Answer 16

You can solve for u and find du that way or do it that way. That way is much more complicated^^.

Answer 17

both ways shown wud b gr8!

Answer 18

After doing long division the integral gets simplified to this:
\[\int\limits u^2+u+1+\frac{1}{u-1} du\]
Integrate term by term giving:
\[\frac{u^3}{3}+\frac{u^2}{2}+u+\ln|u-1|+C\]
Then plug back in. U=x^(1/3) so this becomes:
\[\frac{x}{3}+\frac{1}{2}x^{2/3}+x^{1/3}+\ln|x^{1/3}-1|+C\]

Answer 19

Then of course, all of that its multiplied by 3 (I forgot to copy it).

Answer 20

Wolfram concurs.

Answer 21

ohhhh...got it! thank you so much! i wish my teacher explained them well!:

Answer 22

Youre welcome :D You can post as many as you need help with. I love calc haha

Answer 23

I would probably prefer a u-sub of \[u = x^{1/3} - 1\]\[ \implies du = {1 \over 3x^{2/3}}dx\]\[\implies dx = 3(u+1)^2du\]\[\implies \int {x^{1/3}\over x^{1/3} - 1}dx = 3\int {(u+1) \over u} (u+1)^2 du = 3\int {(u+1)^3 \over u} du\]
... etc.

Though I wasn't really that strong at polynomial division when I was doing this stuff and I'd probably try your way instead.

Answer 24

Now that I'm much better at it I mean.

Answer 25

I know polpak :DDD haha

Answer 26

Thats the nice thing about integrals, they take some thought sometimes xD

Answer 27

haha i hate giving so much thought in math..plug in formulas rock!:) im prob ganna need help with tons more..got my finals on this stuff n derivatives n limits tom!

Answer 28

Just keep posting in the feed, I'll be on for a while xP

Answer 29

Careful though.. the first one is always free.. after that we make you work for it ;p