Q.If $\dfrac{kx}{(x+2)(x-1)}=\dfrac2{x+2}+\dfrac1{x-1}$, then the value of $k$ is

A proper fraction $\dfrac{f(x)}{g(x)}$ ($\deg f<\deg g$) decomposes uniquely into simpler pieces once $g(x)$ is factored into linear and irreducible-quadratic factors:

• each linear factor $(x-a)^k$ contributes $\dfrac{A_1}{x-a}+\dfrac{A_2}{(x-a)^2}+\cdots+\dfrac{A_k}{(x-a)^k}$;
• each irreducible quadratic factor $(x^2+ax+b)^k$ contributes $\dfrac{B_1x+C_1}{x^2+ax+b}+\cdots+\dfrac{B_kx+Ck}{(x^2+ax+b)^k}$.

Finding the constants. Clear denominators, then either substitute each linear factor's root directly (the cover-up rule -- instantly isolates that factor's constant, since every other term vanishes), or -- required whenever a quadratic or repeated factor is present -- expand and match coefficients of like powers of $x$ on both sides; a convenient extra substitution (often $x=0$) frequently speeds up the coefficient-matching step. …