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anonymous
 4 years ago
\[\int_{0}^{1} \frac{1}{1+\sqrt[3]{x} } dx\]
\[u=\sqrt[3]{x}\]
\[du=\frac 1 3 \frac{1}{\sqrt[3]{x^2}}dx\]
\[3du=\frac{1}{\sqrt[3]{x^2}}dx\]
\[3\sqrt[3]{x^2}du=dx\]
\[3u^2du=dx \*\possible\error\]
\[\int_{0}^{1} \frac{3u^2}{1+u } du\]
\[3\int_{0}^{1} \frac{u^2}{1+u } du\]
long division
\[u1+\frac{1}{u+1}\]
\[3\int_{0}^{1} udu3\int_{0}^{1} 1du+3\int_{0}^{1} \frac{1}{u+1}du\]
something is missing...
<looking sideways :\ >
anonymous
 4 years ago
\[\int_{0}^{1} \frac{1}{1+\sqrt[3]{x} } dx\] \[u=\sqrt[3]{x}\] \[du=\frac 1 3 \frac{1}{\sqrt[3]{x^2}}dx\] \[3du=\frac{1}{\sqrt[3]{x^2}}dx\] \[3\sqrt[3]{x^2}du=dx\] \[3u^2du=dx \*\possible\error\] \[\int_{0}^{1} \frac{3u^2}{1+u } du\] \[3\int_{0}^{1} \frac{u^2}{1+u } du\] long division \[u1+\frac{1}{u+1}\] \[3\int_{0}^{1} udu3\int_{0}^{1} 1du+3\int_{0}^{1} \frac{1}{u+1}du\] something is missing... <looking sideways :\ >

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where did the crowd go? Come back....!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not very good at these I'm afraid.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes you are! believe in yourself!1 We'll work through this together

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1busy reading through it now.. how do you know you are wrong btw?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well...I simply slapped a 2 over the u. But It didn't seem to belong there since it was originally under the radicant

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1it's correct. Also, not sure what you mean by "slapped a 2 over the u" haha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i throw my numbers around...ok well, what about the long division?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0silly question, what do I have to do to mathematically check that....what do I multiply it to? I know, silly question....

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1should give you the same thing you started with

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aaahhhhh...got it! create a common denominator and add

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1haha yeah, that's it. sorry, I'm not very clear haha

slaaibak
 4 years ago
Best ResponseYou've already chosen the best response.1glad to help, although you mastered it yourself :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep... i don't see anything wrong...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks everyone! Just needed the reassurance. I don't trust my own book keeping
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