After long-division and factoring the denominator, we have

The partial fraction decomposition takes the form

Multiplying through by ("x" − 1)("x" + 1) we have the polynomial identity

Taking "x" = 1 gives 4 = 4"C", so "C" = 1. Similarly, taking "x" = "i" gives 2 + 2"i" = ("Fi" + "G")(2 + 2"i"), so "Fi" + "G" = 1, so "F" = 0 and "G" = 1 by equating real and imaginary parts. With "C" = "G" = 1 and "F" = 0, taking "x" = 0 we get "A" − "B" + 1 − "E" − 1 = 0, thus "E" = "A" − "B".

We now have the identity

Expanding and sorting by exponents of x we get

We can now compare the coefficients and see that

with "A" = 2 − "D" and −"A" −3 "D" =−4 we get "A" = "D" = 1 and so "B" = 0, furthermore is "C" = 1, "E" = "A" − "B" = 1, "F" = 0 and "G" = 1.

The partial fraction decomposition of "ƒ"("x") is thus

Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at "x=1" and at "x=i" in the above polynomial identity. (To this end, recall that the derivative at "x=a" of "(x−a)p(x)" vanishes if "m > 1" and it is just "p(a)" if "m=1".)
Thus, for instance the first derivative at "x=1" gives 
that is "8 = 4B + 8" so "B=0".

उदाहरण ५ (limit method)
उदाहरण २
उदाहरण ३
विधि