Partial Fractions- Square on Bottom?

Question

Hey i am looking at this partial fraction example and struggling to fathom it (i am a bit rusty) any help would be appreciated.

(x^2-1)/(x)*(x^2+1) = A/x + B*x+C/(x^2+1)

I know the answer just can't figure out this step.
Thanks.

Answer 1

the name i learned for this is fraction splitting

you can split into A/x + (Bx+c)/(x^2 + 1)

( because x^2+1 is irreducible
for x = 1 and -1 the fractions should be zero that speeds up the calculation od A,B C,
finally write out and eqalt the corresponding fractions

Answer 2

multiply by x(x^2+1) to get

x^2-1 = A(x^2+1) +(Bx+C) x

above is identity
constant term on both sides
gives A = -1

coefficient of x on both sides gives C = 0
coefficient of x^2
1 = A + B so B = 2

we get -1/x + 2x/(x^2+1)

check = (-x^2-1+2x^2)/x(x^2+1) or (x^2-1)/(x(x^2+1))

Answer 3

Have this as a general rule.. If u hav a square in the denominator, tak two variables with an one multiplied with x.. (eg:Bx+C, as u hav x^2 in d denominator.. ) And hence the given is split into 3 parts.. A/x +(Bx+C)/(x^2+1).. hope u got it...