Decomposition by Partial Fractions

When the integrand is a proper rational (algebraic) fraction $\dfrac{p(x)}{q(x)}$ — a ratio of polynomials with $q(x)\ne0$ and $\deg p(x)<\deg q(x)$ — and it does not simplify by the direct decomposition tricks of §11.7.1, the standard route is to rewrite it as a sum of simpler fractions with the same denominator factors: partial fractions. If $\deg p(x)\ge\deg q(x)$, perform polynomial long division first, so the improper fraction becomes (polynomial) $+$ (a genuinely proper fraction), and apply partial fractions only to the proper remainder.

How the decomposition is set up, factor by factor of $q(x)$:

• a non-repeated linear factor $(x-a)$ contributes a term $\dfrac{A}{x-a}$;
• a repeated linear factor $(x-a)^k$ contributes $\dfrac{A_1}{x-a}+\dfrac{A_2}{(x-a)^2}+\cdots+\dfrac{A_k}{(x-a)^k}$;
• an irreducible quadratic factor $(x^2+bx+c)$ (no real roots) contributes a term with a linear numerator, $\dfrac{Bx+C}{x^2+bx+c}$.

The unknown constants are found by clearing denominators (multiplying both sides by $q(x)$) and then either substituting convenient values of $x$ (in particular, $x=$ each root of a linear factor makes every other term vanish and pins that constant down immediately) or comparing coefficients of matching powers of $x$ on both sides — usually a mix of both is fastest.

Two worked-style illustrations of the technique (paralleling Example 11.29). For $\dfrac{3x+7}{x^2-3x+2}=\dfrac{3x+7}{(x-2)(x-1)}$, write $\dfrac{3x+7}{(x-2)(x-1)}=\dfrac{A}{x-2}+\dfrac{B}{x-1}$; clearing denominators gives $3x+7=A(x-1)+B(x-2)$, and substituting $x=2$ gives $A=13$, $x=1$ gives $B=-10$, so the integral is $13\log|x-2|-10\log|x-1|+c$. For a denominator with a repeated factor, $\dfrac{x+3}{(x+2)^2(x+1)}=\dfrac{A}{x+2}+\dfrac{B}{(x+2)^2}+\dfrac{C}{x+1}$; clearing denominators and substituting $x=-2$ and $x=-1$ pins down $B$ and $C$ directly, and comparing the coefficient of $x^2$ (which must vanish, since the left side has none) pins down $A$; each resulting piece is then either a $\log$ (for a first-power denominator) or a negative power (for $(x+2)^{-2}$, integrating to $-\dfrac{1}{x+2}$). …