integrate dy/(1-y^2)

Question

hartnn · Answer

do you know about partial fractions ?

amistre64 · Answer

1+y^2+y^6 .....
           -----------------
1 - y^2 | 1
             (1 - y^2)
                   y^2
                   (y^2 - y^4)
                             y^4
                            (y^4 - y^6)
                                      ....

$$\int~\sum y^{2n}~dy=\sum\frac{1}{2n+1}y^{2n+1}$$

amistre64 · Answer

2n is bad ... got confused on the 2468 parts poping out

hartnn · Answer

i am so lost lol
whay is all that needed ?

amistre64 · Answer

its not needed, but it is a way to approach it :)

amistre64 · Answer

the long division process creates a polynomial representation of it; which can be integrated by the power rule

hartnn · Answer

1+y^2+y^6 .... is not y^2n ?

amistre64 · Answer

yeah, i caught that afterwards :)  needed more MtDew

hartnn · Answer

but its always good to know there ia another way to solve any problem :) thanks!

amistre64 · Answer

i was antisipating the y^6 that was coming up
                               vv
               1+y^2+y^4 .....
           -----------------
1 - y^2 | 1
             (1 - y^2)
                   y^2
                   (y^2 - y^4)
                             y^4
                            (y^4 - y^6)

amistre64 · Answer

its the summation representation of a hyperbolic function, which can also be written in log format