$$\frac{x+1}{(x-1)(x-3)}+\frac{x-2}{(x+2)(x-3)} = 2-\frac{2x+3}{x-3}$$
Answer
$$ \begin{matrix}x_1 = 0.13622+0.81248i & x_2 = 0.13622-0.81248i & x_3 = 2.86378+2.24937i \\[1 em] x_4 = 2.86378-2.24937i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x+1}{(x-1)(x-3)}+\frac{x-2}{(x+2)(x-3)} &= 2-\frac{2x+3}{x-3}&& \text{multiply ALL terms by } \color{blue}{ (x-1)(x-3)(x+2) }. \\[1 em](x-1)(x-3)(x+2)\frac{x+1}{(x-1)(x-3)}+(x-1)(x-3)(x+2)\frac{x-2}{(x+2)(x-3)} &= (x-1)(x-3)(x+2)\cdot2-(x-1)(x-3)(x+2)\frac{2x+3}{x-3}&& \text{cancel out the denominators} \\[1 em]x^4-3x^3-7x^2+15x+18+x^4-9x^3+29x^2-39x+18 &= 2x^3-4x^2-10x+12-(2x^3+5x^2-x-6)&& \text{simplify left and right hand side} \\[1 em]2x^4-12x^3+22x^2-24x+36 &= 2x^3-4x^2-10x+12-2x^3-5x^2+x+6&& \\[1 em]2x^4-12x^3+22x^2-24x+36 &= 2x^3-4x^2-10x+12-2x^3-5x^2+x+6&& \\[1 em]2x^4-12x^3+22x^2-24x+36 &= -9x^2-9x+18&& \text{move all terms to the left hand side } \\[1 em]2x^4-12x^3+22x^2-24x+36+9x^2+9x-18 &= 0&& \text{simplify left side} \\[1 em]2x^4-12x^3+31x^2-15x+18 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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