Integrating Using Linear Partial Fractions BC only
A proper rational function whose denominator factors into distinct linear factors can be written with one unknown constant over each factor, as in $\frac{1}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$. Those constants are the whole content of the method: $\frac{1}{x-3} + \frac{1}{x+2}$ is a different function entirely, recombining to $\frac{2x-1}{(x-3)(x+2)}$. Clearing denominators gives $1 = A(x+2) + B(x-3)$, and substituting each root kills one term at a time, giving $A = \frac{1}{5}$ and $B = -\frac{1}{5}$. Constants carry signs, and recombining the pieces is a fifteen-second check that catches every sign error.
Each piece integrates to a logarithm, $\int \frac{k}{x-r}dx = k\ln|x-r| + C$, with the absolute value bars kept and one constant at the end. If the fraction is improper it must be divided first, as in Topic 6.10, and if the denominator does not factor the tool is completing the square instead. For a definite integral, check before anything else that no root of the denominator lies inside the interval: if one does, the integrand blows up there and the integral is improper, which is Topic 6.13's subject rather than this one's.
The work
3 ways in · any order
Lesson
Integrating Using Linear Partial Fractions
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Checks the two conditions a decomposition needs, sets one unknown constant over each distinct linear factor, clears denominators and solves by substituting each root in turn, integrates the pieces into logarithms with their bars and signs intact, and checks for denominator roots inside the interval before evaluating anything.
Diagnostic
10-item topic check
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Ten items on decomposing and integrating rational functions with distinct linear factors: the wrong factor set, numerators of the wrong form, constants never solved for or solved with the wrong sign, and decompositions attempted on fractions that were never made proper.
Targeted Practice
Drill a single misconception
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Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.