Originally Posted by

**earboth** Hello,

I tried this way:

1rst step:

$\displaystyle {2a^2\over{s^4-a^4}}={A\over{a^2+s^2}}+{B\over{a^2-s^2}}+{C\over{s^4-a^4}}$

Then you'll get the results: A = B = 1, C = 0

2nd step (Take the second summand to partial fractions):

$\displaystyle {1\over{a^2-s^2}}={A\over{a+s}}+{B\over{a-s}}+{C\over{a^2-s^2}}$

Then you'll get the results: A = B = 0, C = 1

Thus the only partial fraction I got is:

$\displaystyle {2a^2\over{s^4-a^4}}={1\over{a^2+s^2}}+{1\over{a^2-s^2}}$

Is this sufficient?

Greetings

EB