Integrating Using Linear Partial Fractions

A fraction like 1/((x-1)(x+2)) has no obvious antiderivative as written, but every log rule you know works on something shaped like 1/(x-a). Partial fractions exploit this by rewriting the ugly fraction as A/(x-1) + B/(x+2) for constants A and B. Each piece is now a simple log integral, so the sum of two easy antiderivatives replaces one hard one. This only works when the denominator factors into distinct linear pieces, because that factoring is what tells you which simple fractions to build.

To find A and B, multiply both sides by the full denominator (x-1)(x+2) and clear fractions, giving an equation like 1 = A(x+2) + B(x-1). Plug in x = 1 to kill the B term and solve for A, then plug in x = -2 to kill the A term and solve for B. This shortcut, called cover-up, works because those x-values are exactly the roots of the original denominator. When the numerator is linear, like 3x+11 over (x-3)(x+2), the same cover-up trick still finds both constants in one pass.

Partial fractions only handles denominators that factor into linear pieces over the real numbers. An irreducible quadratic factor, like x²+1, has no real root to plug in, so cover-up fails and a different setup (numerator Cx+D over that quadratic) is needed instead, which sits just outside this method's scope. This technique becomes essential when solving the logistic differential equation dP/dt = kP(1 - P/L), where separating variables leaves an integral of the form 1/(P(L-P)) dP that only partial fractions can crack open into the logistic function's closed form.