Part (1)

10x2-2x+4 x3 + x = A x + Bx+C x2+1 .

In order to work out the constants A,B,C, I put the RHS of the identity over a common denominator:

10x2-2x+4 x3 + x = A(x2+1)+ x(Bx+C) x(x2+1) .

Since the denominators are equal, it follows that the numerators of the fraction must also be equal:

10x2-2x+4 = A(x2+1)+ x(Bx+C) = Ax2+Bx2+Cx+A = (A+B)x2+Cx+A .

By comparing coefficients, it follows that A+B=10, C=-2 and A=4 so B=6.

Part (2)

A x + Bx+C x2+1 = A x + D (x-i) + E (x+i) .

By taking A/x away and replacing B and C by their respective values, and putting D and E over a common denominator:

6x-2 (x2+1) = D(x+i)+E(x-i) (x-i)(x+i) .

Again, since the denominators are equal, it follows that the numerators are equal so 6x-2=D(x+i)+E(x-i).

[Editor's note: We need to take care here as the identity is not defined forx=±i or x=o (when the denominators of the fractions are zero).]

By comparing coefficients we have D+E=6 and (D-E)i=-2. Then 6i-2=D(2i) so D=(6i-2)/2i. Multiply both the numerator and the denominator by i to get a real denominator:

D= 6i2-2i -2 = -6-2i -2 = 3+i.

Then E(-2i) = -6i-2, so E=3-i. So A=4, B=6, C=-2, D=3+i and E=3-i.

10x2-2x+4 x3 + x = 4 x + 6x-2 x2+1 = 4 x + 3+i (x-i) + 3-i (x+i) .

[Editor's note: To be absolutely rigorous, since the expression is undefined for certain values of x, we should consider limits and we get the same result:

D = lim x® i  6x-2 x+i = 6i-2 2i = 3+i.

and similarly

E = lim x® -i  6x-2 x-i = -6i-2 -2i = 3-i.]