Question

solve this integral

Answer 1

\[\int\limits_{0}^{1} \frac{ 1 }{ y^3+1 }\]

Answer 2

i tried to do partial fractions knowing that the denominator is a^3 + b^3 = (a-b)(a^2-ab+b^2)

Answer 3

yes that is what you need
it is real ugly

Answer 4

part of it will be \[\frac{1}{3}\int\frac{dy}{y+1}\]

Answer 5

I think one of your signs is off.
This is the sum of cubes formula:
\(\large\rm a^3+b^3=(a+b)(a^2-ab+b^2)\)

Answer 6

Here is some of the work to get the problem started.
Hopefully it helps get you on the right track.

Answer 7

*golf clap*