6.12 Integrating Using Linear Partial Fractions (BC only)

A rational function whose denominator factors into distinct linear pieces can be rewritten as a sum of simple fractions, each a constant over one linear factor. Every such fraction integrates to a logarithm, so this page starts with that one-term rule and then builds the decomposition on top of it. The technique is BC only.

Recognizing reverse chain-rule structure

Integration methods often begin with pattern recognition. In

$$ \int \frac{A}{ax+b},dx, $$

the denominator $ax+b$ has derivative $a$, a constant factor already present in the numerator after adjustment. Set

$$ u=ax+b, \qquad du=a,dx. $$

Because $dx=\frac{du}{a}$,

$$ \int \frac{A}{ax+b},dx

\frac{A}{a}\int \frac{1}{u},du

\frac{A}{a}\ln|u|+C

\boxed{\frac{A}{a}\ln|ax+b|+C}. $$

For example,

$$ \int \frac{6}{3x-2},dx

2\ln|3x-2|+C. $$

This is a reverse chain-rule move: recognize the derivative of the inner linear expression, adjust by its constant factor, and obtain a logarithm.

Linear partial fractions

A rational function is a quotient of polynomials. When its denominator factors into distinct, nonrepeating linear factors, the rational function may be decomposed into a sum of simpler ratios. This is the central idea of linear partial fractions, identified by FUN-6.F and FUN-6.F.1.

Worked example: decomposition and integration

Decompose

$$ \frac{5x+1}{(x+1)(x+2)}. $$

Because the denominator has two distinct linear factors, write

$$ \frac{5x+1}{(x+1)(x+2)}

\frac{A}{x+1}+\frac{B}{x+2}. $$

Multiplying by $(x+1)(x+2)$ gives

$$ 5x+1=A(x+2)+B(x+1). $$

Expanding and matching coefficients,

$$ 5x+1=(A+B)x+(2A+B), $$

so

$$ A+B=5, \qquad 2A+B=1. $$

Subtracting the first equation from the second gives $A=-4$, and then $B=9$. Thus,

$$ \frac{5x+1}{(x+1)(x+2)}

-\frac{4}{x+1}+\frac{9}{x+2}. $$

Integrate term by term:

$$ \begin{aligned} \int \frac{5x+1}{(x+1)(x+2)},dx &= -4\int\frac{1}{x+1},dx + 9\int\frac{1}{x+2},dx\ &= \boxed{-4\ln|x+1|+9\ln|x+2|+C}. \end{aligned} $$

Each term uses the linear-denominator rule with $a=1$. If the denominator were $4x-3$, the corresponding term would integrate as

$$ \int\frac{A}{4x-3},dx

\frac{A}{4}\ln|4x-3|+C. $$

The two recognition patterns

The methods can be distinguished by the structure of the integrand:

Integrand structure	Recognition cue	Main move
Product such as $x e^x$	Reverse product rule	$\int u,dv=uv-\int v,du$
Rational function with distinct linear factors	Split the denominator into linear pieces	Partial-fraction decomposition
Ratio with a linear denominator	Numerator is related to the denominator’s derivative	Reverse chain rule and logarithm

For FUN-6.F, linear partial fractions are used to determine indefinite or definite integrals of rational functions once the denominator has been factored into distinct linear factors. The associated suggested skill is 1.E — Apply appropriate mathematical rules or procedures, with and without technology.

Retrieval check: For

$$ \int \frac{7}{5x+1},dx, $$

identify the substitution and the coefficient multiplying the logarithm. Then decide whether

$$ \int x\cos x,dx $$

is better suited to substitution or integration by parts. The answers are $u=5x+1$, coefficient $\frac{7}{5}$, and integration by parts.