Integration by Partial Fractions

A rational function $\dfrac{f(x)}{g(x)}$ ($g(x)\neq0$) is called proper if $\deg f<\deg g$, and improper otherwise; an improper fraction is first written as $\text{Quotient}+\dfrac{\text{Remainder}}{g(x)}$ by polynomial division, where the remainder term is proper. A proper rational function is then split into partial fractions according to how $g(x)$ factors — into non-repeated linear factors, a repeated linear factor, or a linear factor times a non-repeated irreducible quadratic (the three forms tabulated above) — and each simpler piece is integrated using the elementary and special-integral formulae already established.

Type (i) — non-repeated linear factors. $\int\dfrac{3x^2+4x-5}{(x^2-1)(x+2)}\,dx$. Factor the denominator fully: $(x-1)(x+1)(x+2)$. Set $\dfrac{3x^2+4x-5}{(x-1)(x+1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{C}{x+2}$; clearing denominators, $3x^2+4x-5=A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)$. Substituting the roots of each factor in turn: at $x=1$, $2=6A\Rightarrow A=\frac13$; at $x=-1$, $-6=-2B\Rightarrow B=3$; at $x=-2$, $-1=3C\Rightarrow C=-\frac13$. Integrating each term: $I=\dfrac13\log|x-1|+3\log|x+1|-\dfrac13\log|x+2|+c=\dfrac13\log\left|\dfrac{(x-1)(x+1)^9}{x+2}\right|+c$.

A variant of Type (i) arises when the denominator's factors are themselves quadratics in disguise: $\int\dfrac{2x^2-3}{(x^2-5)(x^2+4)}\,dx$. Put $m=x^2$, giving the proper rational function $\dfrac{2m-3}{(m-5)(m+4)}=\dfrac{A}{m-5}+\dfrac{B}{m+4}$; solving (at $m=5$: $A=\frac79$; at $m=-4$: $B=\frac{11}9$) and substituting back $m=x^2$: $I=\dfrac79\int\dfrac{dx}{x^2-5}+\dfrac{11}9\int\dfrac{dx}{x^2+4}=\dfrac{7}{18\sqrt5}\log\left|\dfrac{x-\sqrt5}{x+\sqrt5}\right|+\dfrac{11}{18}\tan^{-1}\dfrac x2+c$.

Type (i) also appears after a trigonometric substitution turns a trig integral into a rational one: $\int\dfrac{d\theta}{\sin\theta(3+2\cos\theta)}$. Multiply top and bottom by $\sin\theta$: $\dfrac{\sin\theta}{(1-\cos^2\theta)(3+2\cos\theta)}$; put $t=\cos\theta$ (so $\sin\theta\,d\theta=-dt$), giving $-\displaystyle\int\dfrac{dt}{(1-t)(1+t)(3+2t)}$, three distinct linear factors in $t$; solving for $A,B,C$ and integrating gives $I=\dfrac1{10}\log|1-\cos\theta|-\dfrac12\log|1+\cos\theta|+\dfrac25\log|3+2\cos\theta|+c=\dfrac1{10}\log\dfrac{(1-\cos\theta)(3+2\cos\theta)^4}{(1+\cos\theta)^5}+c$.

Type (ii) — repeated linear factor. $\int\dfrac{d x}{2\cos x+\sin2x}$. Simplify $2\cos x+\sin2x=2\cos x(1+\sin x)$, so the integral is $\frac12\displaystyle\int\dfrac{\cos x}{\cos^2x(1+\sin x)}\,dx$. Put $t=\sin x$: $\frac12\displaystyle\int\dfrac{dt}{(1-t^2)(1+t)}=\frac12\int\dfrac{dt}{(1-t)(1+t)^2}$, a repeated factor at $t=-1$. Setting $\dfrac{1}{(1-t)(1+t)^2}=\dfrac{A}{1-t}+\dfrac{B}{1+t}+\dfrac{C}{(1+t)^2}$ and solving ($A=\frac14,\ B=\frac14,\ C=\frac12$), integrating and substituting back $t=\sin x$ gives $I=\dfrac18\log\left|\dfrac{1+\sin x}{1-\sin x}\right|-\dfrac{1}{1+\sin x}+c$. …