Mistake Master

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.

PARTIAL FRACTIONS: THE CONDITIONS AND THE STEPS. 1. IS THE FRACTION PROPER? IF NOT, DIVIDE FIRST. 2. FACTOR THE DENOMINATOR COMPLETELY. 3. ONE CONSTANT OVER EACH DISTINCT LINEAR FACTOR. 4. CLEAR DENOMINATORS AND SOLVE FOR THE CONSTANTS. 5. INTEGRATE: EACH PIECE IS A LOGARITHM. THE COURSE SCOPE IS DISTINCT LINEAR FACTORS ONLY. 1/(x(x + 2)) IS NOT 1/x + 1/(x + 2). THE CONSTANTS ARE WHAT THE METHOD EXISTS TO FIND. CHECK BY PUTTING THE PIECES BACK OVER ONE DENOMINATOR.
Steps 1 and 2 are the conditions rather than the work. Skipping them is how a decomposition ends up with constants that no substitution can solve.
∫ dx/(x² − x − 6), WORKED. FACTOR: (x − 3)(x + 2) SET UP: A/(x − 3) + B/(x + 2) CLEAR: 1 = A(x + 2) + B(x − 3) PUT x = 3: 1 = 5A, SO A = 1/5 PUT x = −2: 1 = −5B, SO B = −1/5 INTEGRATE EACH PIECE (1/5) ln|x − 3| − (1/5) ln|x + 2| + C CHOOSING x AT EACH ROOT KILLS ONE CONSTANT AT A TIME. THE CONSTANTS CARRY SIGNS. HERE ONE OF THEM IS NEGATIVE. RECOMBINE THE PIECES TO CHECK BEFORE INTEGRATING ANYTHING.
The substitution $x = 3$ is legal even though $x = 3$ is not in the domain of the original fraction, because the cleared equation $1 = A(x+2) + B(x-3)$ is a polynomial identity that holds everywhere.

The work

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Lesson
Integrating Using Linear Partial Fractions

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.

Skill check · 10 scenarios
Diagnostic
10-item topic check

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.

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Targeted Practice
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