anonymous
  • anonymous
\[\int_{0}^{1} \frac{1}{1+\sqrt[3]{x} } dx\] \[u=\sqrt[3]{x}\] \[du=\frac 1 3 \frac{1}{\sqrt[3]{x^2}}dx\] \[3du=\frac{1}{\sqrt[3]{x^2}}dx\] \[3\sqrt[3]{x^2}du=dx\] \[3u^2du=dx \*\possible\error\] \[\int_{0}^{1} \frac{3u^2}{1+u } du\] \[3\int_{0}^{1} \frac{u^2}{1+u } du\] long division \[u-1+\frac{1}{u+1}\] \[3\int_{0}^{1} udu-3\int_{0}^{1} 1du+3\int_{0}^{1} \frac{1}{u+1}du\] something is missing...
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
where did the crowd go? Come back....!!!!!
anonymous
  • anonymous
I'm not very good at these I'm afraid.
anonymous
  • anonymous
yes you are! believe in yourself!1 We'll work through this together

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slaaibak
  • slaaibak
busy reading through it now.. how do you know you are wrong btw?
anonymous
  • anonymous
well...I simply slapped a 2 over the u. But It didn't seem to belong there since it was originally under the radicant
slaaibak
  • slaaibak
it's correct. Also, not sure what you mean by "slapped a 2 over the u" haha
anonymous
  • anonymous
i throw my numbers around...ok well, what about the long division?
slaaibak
  • slaaibak
also correct
anonymous
  • anonymous
silly question, what do I have to do to mathematically check that....what do I multiply it to? I know, silly question....
slaaibak
  • slaaibak
you just add it up
anonymous
  • anonymous
slaaibak
  • slaaibak
should give you the same thing you started with
anonymous
  • anonymous
aaahhhhh...got it! create a common denominator and add
slaaibak
  • slaaibak
haha yeah, that's it. sorry, I'm not very clear haha
anonymous
  • anonymous
thank you @slaaibak
slaaibak
  • slaaibak
glad to help, although you mastered it yourself :)
anonymous
  • anonymous
yep... i don't see anything wrong...
anonymous
  • anonymous
Thanks everyone! Just needed the reassurance. I don't trust my own book keeping

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