Rational Functions and Partial Fractions

The core forms the bank draws on:

• $\displaystyle\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$
• $\displaystyle\int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| + C$
• $\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a} + C$
• $\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + C$
• $\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C$
• $\displaystyle\int \dfrac{dx}{x\sqrt{x^2 - a^2}} = \dfrac{1}{a}\sec^{-1}\dfrac{x}{a} + C$

One companion reflex from substitution closes the set: $\displaystyle\int \dfrac{f'(x)}{\sqrt{f(x)}}\,dx = 2\sqrt{f(x)} + C$ — the $\sqrt{}$ term that appears whenever a numerator is the derivative of what's under the root. Quadratic-in-$x^2$ denominators often reduce to the arctan form via a $t = x \pm \dfrac{k}{x}$ substitution (taught in Substitution → reciprocal substitutions).

Rewrite $ax^2 + bx + c$ as $a\!\left(x + \dfrac{b}{2a}\right)^2 + \left(c - \dfrac{b^2}{4a}\right)$. Then $\displaystyle\int \dfrac{dx}{\sqrt{k^2 - (x+p)^2}} = \sin^{-1}\dfrac{x+p}{k}$, and $\displaystyle\int \dfrac{dx}{(x+p)^2 + k^2} = \dfrac{1}{k}\tan^{-1}\dfrac{x+p}{k}$.

For $\displaystyle\int \dfrac{px+q}{\sqrt{ax^2+bx+c}}\,dx$ (or with the quadratic itself in the denominator), write $px + q = A\dfrac{d}{dx}(ax^2+bx+c) + B$. Match coefficients to find $A, B$. Then $\displaystyle\int \dfrac{A\,(ax^2+bx+c)'}{\sqrt{ax^2+bx+c}}\,dx = 2A\sqrt{ax^2+bx+c}$, and the leftover $\displaystyle B\!\int \dfrac{dx}{\sqrt{ax^2+bx+c}}$ is finished by completing the square.

For a proper rational function, decompose by the denominator's factors:

• distinct linear $(x-a)(x-b)$: $\dfrac{A}{x-a} + \dfrac{B}{x-b}$
• repeated linear $(x-a)^2$: $\dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}$
• irreducible quadratic $(x^2 + c)$: $\dfrac{Ax + B}{x^2 + c}$

Find the constants by the cover-up method or by equating coefficients; integrate each piece.