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Calculus BC · Unit 6 · 6.12
AP Calculus BC · Integration and Accumulation of Change
Integrating Using Linear Partial Fractions
Split one hard fraction into two easy ones — each of which integrates to a logarithm.
key idea$\frac{N(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}$
The procedure
A single fraction with a factorable denominator can be split into simpler fractions, each of which integrates to a logarithm. The AP scope is deliberately narrow: distinct linear factors only, with the numerator's degree lower than the denominator's.
1
Check the degree
The numerator's degree must be less than the denominator's. If it is not, do polynomial long division first and apply partial fractions to the remainder.
in a rational function out a proper fraction
2
Factor the denominator
Into distinct linear factors: $(x-a)(x-b)$. Repeated or irreducible quadratic factors are outside the AP scope.
out the factors
3
Write the decomposition with unknowns
One term per factor: $\dfrac{A}{x-a}+\dfrac{B}{x-b}$. Each numerator is a constant because each denominator is linear.
out a template with $A$ and $B$
4
Clear denominators and solve
Multiply through by the full denominator, then substitute the roots — each root kills one unknown and reveals the other immediately.
out the numeric values of $A$ and $B$
5
Integrate each simple fraction
$\displaystyle\int\dfrac{A}{x-a}\,dx=A\ln\lvert x-a\rvert$. Every term becomes a logarithm.
out the answer
| Denominator | In AP scope? | Decomposition |
|---|
| $(x-1)(x+3)$ | yes | $\dfrac{A}{x-1}+\dfrac{B}{x+3}$ |
| $x^{2}-4$ | yes | factors as $(x-2)(x+2)$ |
| $x(x-5)$ | yes | $\dfrac{A}{x}+\dfrac{B}{x-5}$ |
| $(x-2)^{2}$ | no | repeated factor — beyond AP |
| $x^{2}+1$ | no | irreducible quadratic — beyond AP |
If the denominator does not split into distinct linear factors, the AP course does not expect partial fractions. Look for substitution instead.
Every termbecomes a logarithm — $\ln\lvert x-a\rvert$
Solve fastsubstitute each root; it kills all but one unknown
AP scopedistinct linear factors only
Worked examples
EX 1
The standard decomposition
$\int\dfrac{5x-1}{x^{2}-x-2}\,dx$Evaluate $\displaystyle\int\dfrac{5x-1}{x^{2}-x-2}\,dx$.
1
Check degrees and factor
Numerator degree $1$, denominator degree $2$ — proper. Factor: $x^{2}-x-2=(x-2)(x+1)$.
2
Write the template
$\dfrac{5x-1}{(x-2)(x+1)}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$.
3
Clear denominators
$5x-1=A(x+1)+B(x-2)$.
4
Substitute the roots
$x=2$: $9=3A$, so $A=3$. $x=-1$: $-6=-3B$, so $B=2$.
5
Integrate
$\displaystyle\int\left(\dfrac{3}{x-2}+\dfrac{2}{x+1}\right)dx=3\ln\lvert x-2\rvert+2\ln\lvert x+1\rvert+C$.
EX 2
Divide first
improper fractionEvaluate $\displaystyle\int\dfrac{x^{2}+3x}{x^{2}-4}\,dx$.
1
Compare degrees
Both are degree $2$, so the fraction is improper. Partial fractions cannot be applied yet.
2
Divide
$\dfrac{x^{2}+3x}{x^{2}-4}=1+\dfrac{3x+4}{x^{2}-4}$.
3
Decompose the remainder
$\dfrac{3x+4}{(x-2)(x+2)}=\dfrac{A}{x-2}+\dfrac{B}{x+2}$. At $x=2$: $10=4A$, so $A=\tfrac52$. At $x=-2$: $-2=-4B$, so $B=\tfrac12$.
4
Integrate all three pieces
$\displaystyle\int 1\,dx+\int\dfrac{5/2}{x-2}dx+\int\dfrac{1/2}{x+2}dx$.
5
Finish
$x+\dfrac{5}{2}\ln\lvert x-2\rvert+\dfrac{1}{2}\ln\lvert x+2\rvert+C$.
EX 3
The logistic form
$\int\dfrac{dx}{x(1-x)}$Evaluate $\displaystyle\int\dfrac{dx}{x(1-x)}$ — the integral behind the logistic model.
1
Write the template
$\dfrac{1}{x(1-x)}=\dfrac{A}{x}+\dfrac{B}{1-x}$.
2
Clear denominators
$1=A(1-x)+Bx$.
3
Substitute the roots
$x=0$: $1=A$. $x=1$: $1=B$.
4
Integrate
$\displaystyle\int\dfrac{dx}{x}+\int\dfrac{dx}{1-x}=\ln\lvert x\rvert-\ln\lvert 1-x\rvert+C$.
5
Mind the inner sign
$\displaystyle\int\dfrac{dx}{1-x}=-\ln\lvert1-x\rvert$ because the inner derivative is $-1$. Combining gives $\ln\left\lvert\dfrac{x}{1-x}\right\rvert+C$.
EX 4
Definite integral with partial fractions
$\int_3^4$Evaluate $\displaystyle\int_{3}^{4}\dfrac{dx}{x^{2}-x-2}$.
1
Check continuity on the interval
The denominator $(x-2)(x+1)$ vanishes at $2$ and $-1$ — neither is in $[3,4]$, so the FTC applies.
2
Decompose
$\dfrac{1}{(x-2)(x+1)}=\dfrac{1/3}{x-2}-\dfrac{1/3}{x+1}$, from $x=2$ giving $1=3A$ and $x=-1$ giving $1=-3B$.
3
Antidifferentiate
$\dfrac13\ln\lvert x-2\rvert-\dfrac13\ln\lvert x+1\rvert=\dfrac13\ln\left\lvert\dfrac{x-2}{x+1}\right\rvert$.
4
Evaluate
$\dfrac13\left[\ln\dfrac{2}{5}-\ln\dfrac{1}{4}\right]=\dfrac13\ln\dfrac{8}{5}$.
5
Why step 1 mattered
Had the interval straddled $x=2$, the integral would be improper and this computation invalid.
Where it goes wrong
Check the degree before decomposing
An improper fraction must be divided first. Applying partial fractions directly gives an inconsistent system with no solution.
Substitute the roots to solve fast
Setting $x$ to each root kills all but one unknown. Expanding and matching coefficients works too, but takes far longer.
Watch the inner sign on $1-x$
$\int\dfrac{dx}{1-x}=-\ln\lvert1-x\rvert$, not $+$. The inner derivative is $-1$ and it must appear.
Keep the absolute values
$\ln\lvert x-a\rvert$. Dropping the bars makes the antiderivative wrong wherever the argument is negative.