Hey ! what is the integral of x^(1/3)/x^(1/3)-1
substitute x^1/3 = u
\[\int\limits_{}^{}\frac u {u-1}\]
\[\int\limits_{?}^{?} \sqrt[3]{x}\div \sqrt[3]{x}-1\]
yeah
how would u find the deriv of u over u = or - n
use the quotient rule or something. I don't remember, I did this in high school
let me find it and paste it here
alrighty..cool:)
u/(u-1) =1 + 1/(u-1)
ohh...how did u get tht?
there is a technique for it but I dont know that, I just always look at it and think :-)
\[\int\limits \frac{x^{1/3}}{x^{1/3}-1}dx\] 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.
Let me know if you want me to take it further.
take it further plz:)
u=x^(1/3) du=1/3 x^(-2/3) dx isnt this the next step?
You can solve for u and find du that way or do it that way. That way is much more complicated^^.
both ways shown wud b gr8!
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: \[\frac{x}{3}+\frac{1}{2}x^{2/3}+x^{1/3}+\ln|x^{1/3}-1|+C\]
Then of course, all of that its multiplied by 3 (I forgot to copy it).
Wolfram concurs.
ohhhh...got it! thank you so much! i wish my teacher explained them well!:
Youre welcome :D You can post as many as you need help with. I love calc haha
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.
Now that I'm much better at it I mean.
I know polpak :DDD haha
Thats the nice thing about integrals, they take some thought sometimes xD
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!
Just keep posting in the feed, I'll be on for a while xP
Careful though.. the first one is always free.. after that we make you work for it ;p
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