Question

\[\int_{0}^{1} \frac{1}{1+\sqrt[3]{x} } dx\]
\[u=\sqrt[3]{x}\]
\[du=\frac 1 3 \frac{1}{\sqrt[3]{x^2}}dx\]
\[3du=\frac{1}{\sqrt[3]{x^2}}dx\]
\[3\sqrt[3]{x^2}du=dx\]
\[3u^2du=dx \*\possible\error\]
\[\int_{0}^{1} \frac{3u^2}{1+u } du\]
\[3\int_{0}^{1} \frac{u^2}{1+u } du\]
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...

Answer 1

where did the crowd go? Come back....!!!!!

Answer 2

I'm not very good at these I'm afraid.

Answer 3

yes you are! believe in yourself!1 We'll work through this together

Answer 4

busy reading through it now.. how do you know you are wrong btw?

Answer 5

well...I simply slapped a 2 over the u. But It didn't seem to belong there since it was originally under the radicant

Answer 6

it's correct. Also, not sure what you mean by "slapped a 2 over the u" haha

Answer 7

i throw my numbers around...ok well, what about the long division?

Answer 8

also correct

Answer 9

silly question, what do I have to do to mathematically check that....what do I multiply it to?
I know, silly question....

Answer 10

you just add it up

Answer 11

<looking sideways>

Answer 12

should give you the same thing you started with

Answer 13

aaahhhhh...got it! create a common denominator and add

Answer 14

haha yeah, that's it. sorry, I'm not very clear haha

Answer 15

thank you @slaaibak

Answer 16

glad to help, although you mastered it yourself :)

Answer 17

yep... i don't see anything wrong...

Answer 18

Thanks everyone! Just needed the reassurance. I don't trust my own book keeping