4/(2-3x) - 1/(1+x) + 3/(1+x2)

$$\frac{4}{(2-3x)}-\frac{1}{(1+x)}+\frac{3}{(1+x^2)}\\\\
=\frac{4(1+x)(1+x^2)}{(2-3x)(1+x)(1+x^2)}-\frac{1(2-3x)(1+x^2)}{(1+x)(2-3x)(1+x^2)}+\frac{3(1+x)(2-3x)}{(1+x^2)(1+x)(2-3x)}\\\\
=\frac{4(x^3+x^2+x+1)}{(2-3x)(1+x)(1+x^2)}-\frac{1(-3x^3+2x^2-3x+2)}{(1+x)(2-3x)(1+x^2)}+\frac{3(-3x^2-x+2)}{(1+x^2)(1+x)(2-3x)}\\\\
=\frac{(4x^3+4x^2+4x+4)}{(2-3x)(1+x)(1+x^2)}+\frac{(3x^3-2x^2+3x-2)}{(1+x)(2-3x)(1+x^2)}+\frac{(-9x^2-3x+6)}{(1+x^2)(1+x)(2-3x)}\\\\
=\frac{(4x^3+4x^2+4x+4)+(3x^3-2x^2+3x-2)+(-9x^2-3x+6)}{(2-3x)(1+x)(1+x^2)}\\\\
=\frac{7x^3-7x^2+4x+8}{(x^2+1)(-3x+2)(x+1)}\\\\$$

I think that is ok.  It hasn't simplified very nicely - maybe I made an error somewhere