Topic 6.12 Using Linear Partial Fractions: rewrite a rational function with distinct linear factors in the denominator as a sum of partial fractions and integrate each to a logarithm (BC).

What this topic is asking

The College Board (Topic 6.12, BC only) covers integrating a rational function whose denominator factors into distinct linear factors. You cannot integrate $\frac{1}{x(x-1)}$ directly, but you can split it into a sum of simpler fractions, each of which integrates to a logarithm. The technique is called partial fraction decomposition.

The method for distinct linear factors

Making the fraction proper first

Partial fractions requires the numerator degree to be less than the denominator degree. If it is not, you must divide first. For $\frac{x^2}{x^2 - 1}$, long division gives $1 + \frac{1}{x^2 - 1}$, and only the remainder fraction $\frac{1}{x^2 - 1}$ is then decomposed. Skipping this check is a common error: applying the partial-fraction template to an improper fraction produces an inconsistent system with no solution. Always confirm $\deg P < \deg Q$, and divide if it is not.

A worked decomposition and integration

The cover-up shortcut

The substitution step in the worked example is the cover-up method in disguise: to find the constant over $(x - r)$, mentally cover that factor in the original and evaluate the rest at $x = r$. For $\frac{3x+1}{(x+2)(x-1)}$, the constant over $(x-1)$ is $\frac{3(1)+1}{1+2} = \frac{4}{3}$, matching $B$ above. This is faster than setting up the full equation and is reliable whenever the factors are distinct and linear. It works because substituting $x = r$ annihilates every other term, leaving only the constant you want times a known number.

Why every answer is a sum of logarithms

Each partial fraction has the form $\frac{A}{x - r}$, whose antiderivative is $A\ln|x - r|$ by the basic reciprocal rule (a substitution $u = x - r$). So a distinct-linear-factor integral always produces a linear combination of logarithms. This is worth knowing as a sanity check: if your answer to a Topic 6.12 integral is not a sum of $\ln$ terms, something has gone wrong in the decomposition. The logarithms can then be combined using log laws (for example $\frac{5}{3}\ln|x+2| + \frac{4}{3}\ln|x-1|$ stays as is, but $\ln|x+2| + \ln|x-1| = \ln|(x+2)(x-1)|$), though leaving them separated is perfectly acceptable on the exam.