## MathSofiya Group Title $\int_{0}^{1} \frac{1}{1+\sqrt[3]{x} } dx$ $u=\sqrt[3]{x}$ $du=\frac 1 3 \frac{1}{\sqrt[3]{x^2}}dx$ $3du=\frac{1}{\sqrt[3]{x^2}}dx$ $3\sqrt[3]{x^2}du=dx$ $3u^2du=dx \*\possible\error$ $\int_{0}^{1} \frac{3u^2}{1+u } du$ $3\int_{0}^{1} \frac{u^2}{1+u } du$ 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 :\ > 2 years ago 2 years ago

1. MathSofiya Group Title

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

2. Thomas9 Group Title

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

3. MathSofiya Group Title

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

4. slaaibak Group Title

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

5. MathSofiya Group Title

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 Group Title

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

7. MathSofiya Group Title

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

8. slaaibak Group Title

also correct

9. MathSofiya Group Title

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

10. slaaibak Group Title

you just add it up

11. MathSofiya Group Title

<looking sideways>

12. slaaibak Group Title

should give you the same thing you started with

13. MathSofiya Group Title

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

14. slaaibak Group Title

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

15. MathSofiya Group Title

thank you @slaaibak

16. slaaibak Group Title

glad to help, although you mastered it yourself :)

17. dpaInc Group Title

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

18. MathSofiya Group Title

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