Mathematics 85 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

substitute x^1/3 = u

OpenStudy (anonymous):

$\int\limits_{}^{}\frac u {u-1}$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

let me find it and paste it here

OpenStudy (anonymous):

alrighty..cool:)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohh...how did u get tht?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

$\int\limits \frac{x^{1/3}}{x^{1/3}-1}dx$ 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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

take it further plz:)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

both ways shown wud b gr8!

OpenStudy (anonymous):

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: $\frac{x}{3}+\frac{1}{2}x^{2/3}+x^{1/3}+\ln|x^{1/3}-1|+C$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Wolfram concurs.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I know polpak :DDD haha

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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