## 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 $u-1+\frac{1}{u+1}$ $3\int_{0}^{1} udu-3\int_{0}^{1} 1du+3\int_{0}^{1} \frac{1}{u+1}du$ something is missing... <looking sideways :\ >

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. slaaibak

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

5. 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

6. slaaibak

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

7. anonymous

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

8. slaaibak

also correct

9. anonymous

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

10. slaaibak

11. anonymous

<looking sideways>

12. slaaibak

should give you the same thing you started with

13. anonymous

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

14. slaaibak

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

15. anonymous

thank you @slaaibak

16. slaaibak

glad to help, although you mastered it yourself :)

17. anonymous

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

18. anonymous

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

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