Linear Partial Fractions — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

Adding

\frac{2}{x - 1} - \frac{1}{x + 3}

gives

\frac{x + 7}{(x - 1)(x + 3)}

, a fraction no basic rule can integrate. Running the addition backward splits it into two pieces that each integrate to a natural log. That backward step is the method of partial fractions.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $x^2 + x - 6 =$

$(x + 3)(x - 2)$
$(x - 3)(x + 2)$
$(x + 6)(x - 1)$
$(x - 6)(x + 1)$

Show the answer

Two numbers with product $-6$ and sum $1$ : $3$ and $-2$ .

Correct: $(x + 3)(x - 2)$ :
$(x - 3)(x + 2)$ :
$(x + 6)(x - 1)$ :
$(x - 6)(x + 1)$ :

2. $\int \frac{3}{x + 1}\,dx =$

$3\ln|x + 1| + C$
$\frac{3}{(x + 1)^2} + C$
$3(x + 1)^{-2} + C$
$\ln|3x + 3| \cdot 3x + C$

Show the answer

$\int \frac{1}{u}\,du = \ln|u| + C$ , with $u = x + 1$ .

Correct: $3\ln|x + 1| + C$ :
$\frac{3}{(x + 1)^2} + C$ :
$3(x + 1)^{-2} + C$ :
$\ln|3x + 3| \cdot 3x + C$ :

Part 4 · See it

See it first

Graph of y = (x + 7)/((x − 1)(x + 3)), solid violet, with dashed vertical asymptotes at x = 1 and x = −3, and its two pieces dashed: y = 2/(x − 1) in blue and y = −1/(x + 3) in vermillion. At every x the violet curve is the sum of the two pieces, so its integral is 2 ln|x − 1| − ln|x + 3| + C. — The violet curve (x + 7)/((x − 1)(x + 3)) is, at every x, the sum of the blue curve 2/(x − 1) and the vermillion curve −1/(x + 3). Each piece has one asymptote and integrates to a natural log. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The denominator factors into distinct linear factors $(x - a)(x - b)$ the fraction can be written as $\frac{A}{x - a} + \frac{B}{x - b}$
Multiply both sides by $(x - a)(x - b)$ a polynomial identity that holds for every $x$
Substitute $x = a$ , then $x = b$ one term vanishes each time, giving $A$ and then $B$
Each piece has the form $\frac{A}{x - a}$ it integrates to $A\ln|x - a|$
The top's degree is at least the bottom'sdo long division first, then split the remainder

Part 6 · Key ideas

Key ideas

Factor the denominator into distinct linear factors; write one fraction $\frac{A}{x - a}$ per factor.
Clear the denominator and substitute each zero to find the constants. Check by adding the pieces back.
$\int \frac{A}{x - a}\,dx$ $= A\ln|x - a| + C$ ; combine logs only when it helps.
Top degree at least the bottom's: divide first. Top equal to the derivative of the bottom: substitute instead.

Part 7 · Misconception

A common mistake

The wrong idea: $\int \frac{x + 7}{(x - 1)(x + 3)}\,dx$ $= \ln|(x - 1)(x + 3)|$ $+ C$ : the log of the denominator.

What actually happens: The log of the denominator is right only when the top is the derivative of the bottom ( $2x + 2$ here). Split first: $\frac{2}{x - 1} - \frac{1}{x + 3}$ , so the integral is $2\ln|x - 1| - \ln|x + 3| + C$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Splitting $\frac{x + 7}{(x - 1)(x + 3)}$

A student integrates

\frac{x + 7}{(x - 1)(x + 3)}

by linear partial fractions. The table shows the plan.

Steps for the partial fraction decomposition
Step	What the student writes
1. Choose the form	$\frac{x + 7}{(x - 1)(x + 3)}$ $= \frac{A}{x - 1}$ $+ \frac{B}{x + 3}$
2. Clear the denominator	$x + 7$ $= A(x + 3) + B(x - 1)$
3. Substitute the zeros	$x = 1$ , then $x = -3$
4. Integrate each piece	$\int \frac{A}{x - 1}\,dx + \int \frac{B}{x + 3}\,dx$

1. Why does step 3 substitute $x = 1$ and $x = -3$ ?

They are the values where $x + 7$ equals zero
They are the special inputs where the identity in step 2 holds
They make the original fraction equal to zero
Each value makes one term of $A(x + 3) + B(x - 1)$ zero, leaving one unknown

Show the answer

The identity $x + 7$ $= A(x + 3) + B(x - 1)$ holds for every $x$ . Choosing the zeros of the factors makes one term disappear each time.

They are the values where $x + 7$ equals zero: $x + 7 = 0$ at $x = -7$ . The useful values are the zeros of the denominator's factors, which knock out one term each.
They are the special inputs where the identity in step 2 holds: Step 2 holds for every $x$ (it is an identity), which is why any $x$ may be substituted. These two are just the most convenient.
They make the original fraction equal to zero: At $x = 1$ and $x = -3$ the original fraction is undefined. Step 2 has no denominator, so those inputs are fine there.
Correct: Each value makes one term of $A(x + 3) + B(x - 1)$ zero, leaving one unknown: Right: at $x = 1$ the $B$ term vanishes ( $8 = 4A$ ); at $x = -3$ the $A$ term vanishes ( $4 = -4B$ ).

2. Find the value of $A$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

At $x = 1$ : $8 = A(4) + B(0)$ , so $A = 2$ .

Answer: 2

3. Find the value of $B$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

At $x = -3$ : $4 = A(0) + B(-4)$ , so $B = -1$ .

Answer: -1

4. Find $\int \frac{x + 7}{(x - 1)(x + 3)}\,dx$ . Include $+ C$ .

Type an expression for $\int \frac{x + 7}{(x - 1)(x + 3)}\,dx =$ .

Show the answer

$\int \left(\frac{2}{x - 1} - \frac{1}{x + 3}\right)dx$ $= 2\ln|x - 1| - \ln|x$ $+ 3|$ $+ C$ .

Answer: $\int \frac{x + 7}{(x - 1)(x + 3)}\,dx =$ $2\,\ln(\left|x - 1\right|) - \ln(\left|x + 3\right|) + C$

5. Which integral needs long division before linear partial fractions?

$\int \frac{5}{x^2 - 1}\,dx$
$\int \frac{2x}{x^2 - 1}\,dx$
$\int \frac{x^2 + 3}{x^2 - 1}\,dx$
$\int \frac{3x + 1}{x^2 - 1}\,dx$

Show the answer

Partial fractions need the top's degree below the bottom's. When it is not, long division (6.10) comes first.

$\int \frac{5}{x^2 - 1}\,dx$ : A constant top has degree 0, below the bottom's 2: split directly.
$\int \frac{2x}{x^2 - 1}\,dx$ : The top is the derivative of the bottom, so substitution gives $\ln|x^2 - 1| + C$ with no division at all.
Correct: $\int \frac{x^2 + 3}{x^2 - 1}\,dx$ : Right: the top and bottom have the same degree, so divide first: $\frac{x^2 + 3}{x^2 - 1}$ $= 1 + \frac{4}{x^2 - 1}$ .
$\int \frac{3x + 1}{x^2 - 1}\,dx$ : The top has degree 1, below the bottom's 2, so partial fractions apply at once.

6. $\int \frac{1}{(x - 2)(x + 2)}\,dx$ $=$

$\frac{1}{4}\ln\left|\frac{x - 2}{x + 2}\right| + C$
$\ln\left|(x - 2)(x + 2)\right| + C$
$\frac{1}{4}\ln\left|\frac{x + 2}{x - 2}\right| + C$
$\ln|x - 2| - \ln|x + 2| + C$

Show the answer

$1 = A(x + 2) + B(x - 2)$ : $x = 2$ gives $A = \frac{1}{4}$ , $x = -2$ gives $B = -\frac{1}{4}$ .

Correct: $\frac{1}{4}\ln\left|\frac{x - 2}{x + 2}\right| + C$ : Right: $\frac{1}{(x - 2)(x + 2)}$ $= \frac{1/4}{x - 2} - \frac{1/4}{x + 2}$ , and a difference of logs is the log of a quotient.
$\ln\left|(x - 2)(x + 2)\right| + C$ : That would need the top to be the derivative of the bottom, $2x$ . The top is 1, so split the fraction.
$\frac{1}{4}\ln\left|\frac{x + 2}{x - 2}\right| + C$ : The quotient is upside down: $A = \frac{1}{4}$ goes with $x - 2$ , so $x - 2$ is on top.
$\ln|x - 2| - \ln|x + 2| + C$ : The pieces carry $\frac{1}{4}$ : at $x = 2$ , $1 = 4A$ .

Part 9 · Summary

Summary

When a denominator factors into distinct linear factors, split the fraction into

\frac{A}{x - a} + \frac{B}{x - b}

. Clear the denominator, substitute the zeros to find

A

and

B

, and integrate each piece to a natural log. Divide first if the top's degree is not below the bottom's.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Splitting x+7(x−1)(x+3)\frac{x + 7}{(x - 1)(x + 3)}

Summary

What comes next

Connections

Splitting $\frac{x + 7}{(x - 1)(x + 3)}$