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\[\int_{0}^{1} \frac{1}{1+\sqrt[3]{x} } dx\]
\[u=\sqrt[3]{x}\]
\[du=\frac 1 3 \frac{1}{\sqrt[3]{x^2}}dx\]
\[3du=\frac{1}{\sqrt[3]{x^2}}dx\]
\[3\sqrt[3]{x^2}du=dx\]
\[3u^2du=dx \*\possible\error\]
\[\int_{0}^{1} \frac{3u^2}{1+u } du\]
\[3\int_{0}^{1} \frac{u^2}{1+u } du\]
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 :\ >
 one year ago
 one year 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 :\ >
 one year ago
 one year ago

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MathSofiyaBest ResponseYou've already chosen the best response.1
where did the crowd go? Come back....!!!!!
 one year ago

Thomas9Best ResponseYou've already chosen the best response.0
I'm not very good at these I'm afraid.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yes you are! believe in yourself!1 We'll work through this together
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
busy reading through it now.. how do you know you are wrong btw?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
well...I simply slapped a 2 over the u. But It didn't seem to belong there since it was originally under the radicant
 one year ago

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

MathSofiyaBest ResponseYou've already chosen the best response.1
i throw my numbers around...ok well, what about the long division?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
silly question, what do I have to do to mathematically check that....what do I multiply it to? I know, silly question....
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
should give you the same thing you started with
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
aaahhhhh...got it! create a common denominator and add
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
haha yeah, that's it. sorry, I'm not very clear haha
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
thank you @slaaibak
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
glad to help, although you mastered it yourself :)
 one year ago

dpaIncBest ResponseYou've already chosen the best response.0
yep... i don't see anything wrong...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Thanks everyone! Just needed the reassurance. I don't trust my own book keeping
 one year ago
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