Practice / Calculus

Partial fractions and integrating rational functions

Ten rational functions split into partial fractions and integrated: distinct linear factors by the cover-up rule, three factors, a repeated factor, an irreducible quadratic, an improper fraction that needs long division first, and the definite integral of 1/(x² − 1), with worked solutions and the mistakes that choose the wrong form or skip the division.

Before you start

Substitution and parts do not integrate $\tfrac{1}{x^2 - 1}$ but splitting it into $\tfrac12\big(\tfrac{1}{x - 1} - \tfrac{1}{x + 1}\big)$ turns it into two logs. Partial fractions is that split, done systematically: the form of the answer is fixed by the factors of the denominator, and only the constants are unknown. The work goes wrong in the set-up more than in the arithmetic: a repeated or quadratic factor given too few terms, an improper fraction decomposed without dividing first, a cover-up evaluated at the wrong root. These ten problems decompose first, then integrate.

A rational function is $\tfrac{P(x)}{Q(x)}$ with $P$ and $Q$ polynomials. It is proper when $\deg P < \deg Q$ If it is not, divide first: $\tfrac PQ = S + \tfrac RQ$ with $\deg R < \deg Q$ and decompose only $\tfrac RQ$
Factor $Q$ completely into linear factors and irreducible quadratics (quadratics with no real roots, such as $x^2 + 1$ Each factor contributes terms to the form:

Factor of $Q$	Terms in the decomposition
$x - a$ once	$\tfrac{A}{x - a}$
$(x - a)^2$	$\tfrac{A}{x - a} + \tfrac{B}{(x - a)^2}$
$x^2 + px + q$ irreducible	$\tfrac{Bx + C}{x^2 + px + q}$

To find the constants, multiply both sides by $Q$ and either substitute convenient values of $x$ (the roots of $Q$ first) or match coefficients of each power of $x$ The cover-up rule is the substitution done in one step: for a factor $x - a$ that appears once, $A$ is $\tfrac PQ$ with the factor $x - a$ covered up, evaluated at $x = a$ For the factor $x + 2$ $a = -2$
Integrals used: $\int\tfrac{dx}{x - a} = \ln|x - a| + C$ $\int\tfrac{dx}{(x - a)^2} = -\tfrac{1}{x - a} + C$ $\int\tfrac{2x}{x^2 + 1}dx = \ln(x^2 + 1) + C$ $\int\tfrac{dx}{x^2 + 1} = \arctan x + C$

Builds on: Integration: substitution and parts

Problems

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Decompose $\dfrac{5x - 1}{(x + 1)(x - 2)}$ into partial fractions.
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Decompose $\dfrac{x + 7}{x^2 - x - 6}$
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Decompose $\dfrac{2x^2 + 6x - 2}{x(x - 1)(x + 2)}$
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Decompose $\dfrac{x^2 + 3}{x(x - 1)^2}$
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Decompose $\dfrac{2x^2 + 3x + 3}{(x + 1)(x^2 + 1)}$
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Write $\dfrac{x^3 + x^2 + x + 3}{x^2 - 1}$ as a polynomial plus partial fractions.
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Find $\displaystyle\int\frac{5x - 1}{(x + 1)(x - 2)}\,dx$
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Evaluate $\displaystyle\int_2^3\frac{dx}{x^2 - 1}$ exactly.
···
Find $\displaystyle\int\frac{x^2 + 3}{x(x - 1)^2}\,dx$
···
Find $\displaystyle\int\frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 1)}\,dx$

Answers

$\dfrac{2}{x + 1} + \dfrac{3}{x - 2}$
$\dfrac{2}{x - 3} - \dfrac{1}{x + 2}$
$\dfrac{1}{x} + \dfrac{2}{x - 1} - \dfrac{1}{x + 2}$
$\dfrac{3}{x} - \dfrac{2}{x - 1} + \dfrac{4}{(x - 1)^2}$
$\dfrac{1}{x + 1} + \dfrac{x + 2}{x^2 + 1}$
$x + 1 + \dfrac{3}{x - 1} - \dfrac{1}{x + 1}$
$2\ln|x + 1| + 3\ln|x - 2| + C$
$\tfrac12\ln\tfrac32 \approx 0.2027$
$3\ln|x| - 2\ln|x - 1| - \dfrac{4}{x - 1} + C$
$\ln|x + 1| + \tfrac12\ln(x^2 + 1) + 2\arctan x + C$

Worked solutions

Problem 1

Decompose $\dfrac{5x - 1}{(x + 1)(x - 2)}$ into partial fractions.

$\dfrac{5x - 1}{(x + 1)(x - 2)} = \dfrac{A}{x + 1} + \dfrac{B}{x - 2}$ Two distinct linear factors, one constant over each. The fraction is proper: degree 1 over degree 2.
$5x - 1 = A(x - 2) + B(x + 1)$ Multiply both sides by $(x + 1)(x - 2)$ This holds for every $x$
$x = 2$ $9 = 3B$ so $B = 3$ At $x = 2$ the $A$ term is 0. This is the cover-up rule: cover $x - 2$ and evaluate $\tfrac{5x - 1}{x + 1}$ at 2.
$x = -1$ $-6 = -3A$ so $A = 2$ At $x = -1$ the $B$ term is 0.
$\dfrac{2}{x + 1} + \dfrac{3}{x - 2}$ Check: $2(x - 2) + 3(x + 1) = 5x - 1$

Problem 2

Decompose $\dfrac{x + 7}{x^2 - x - 6}$

$x^2 - x - 6 = (x - 3)(x + 2)$ Factor the denominator first: $-3$ and $2$ multiply to $-6$ and add to $-1$
$\dfrac{x + 7}{(x - 3)(x + 2)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 2}$ Two distinct linear factors.
$A = \dfrac{3 + 7}{3 + 2} = 2$ Cover-up for $x - 3$ cover it and evaluate $\tfrac{x + 7}{x + 2}$ at its root, $x = 3$
$B = \dfrac{-2 + 7}{-2 - 3} = \dfrac{5}{-5} = -1$ Cover-up for $x + 2$ its root is $x = -2$ so evaluate $\tfrac{x + 7}{x - 3}$ there.
$\dfrac{2}{x - 3} - \dfrac{1}{x + 2}$ Check: $2(x + 2) - (x - 3) = x + 7$

Problem 3

Decompose $\dfrac{2x^2 + 6x - 2}{x(x - 1)(x + 2)}$

$\dfrac{2x^2 + 6x - 2}{x(x - 1)(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{x + 2}$ Three distinct linear factors; degree 2 over degree 3, so proper.
$A = \dfrac{-2}{(0 - 1)(0 + 2)} = \dfrac{-2}{-2} = 1$ Cover up $x$ and evaluate the rest at $x = 0$
$B = \dfrac{2 + 6 - 2}{1\cdot(1 + 2)} = \dfrac{6}{3} = 2$ Cover up $x - 1$ and evaluate at $x = 1$
$C = \dfrac{8 - 12 - 2}{(-2)(-2 - 1)} = \dfrac{-6}{6} = -1$ Cover up $x + 2$ and evaluate at $x = -2$
$\dfrac{1}{x} + \dfrac{2}{x - 1} - \dfrac{1}{x + 2}$ Check: $(x - 1)(x + 2) + 2x(x + 2) - x(x - 1) = (x^2 + x - 2) + (2x^2 + 4x) - (x^2 - x) = 2x^2 + 6x - 2$

Problem 4

Decompose $\dfrac{x^2 + 3}{x(x - 1)^2}$

$\dfrac{x^2 + 3}{x(x - 1)^2} = \dfrac{A}{x} + \dfrac{B}{x - 1} + \dfrac{C}{(x - 1)^2}$ The repeated factor $(x - 1)^2$ needs a term for each power, $x - 1$ and $(x - 1)^2$ Three unknowns for a cubic denominator.
$x^2 + 3 = A(x - 1)^2 + Bx(x - 1) + Cx$ Multiply both sides by $x(x - 1)^2$
$x = 0$ $3 = A$ $x = 1$ $4 = C$ Each root kills every term but one.
Coefficient of $x^2$ $1 = A + B$ so $B = -2$ No root isolates $B$ so match the $x^2$ terms: $A(x - 1)^2$ gives $Ax^2$ $Bx(x - 1)$ gives $Bx^2$ and $Cx$ has none.
$\dfrac{3}{x} - \dfrac{2}{x - 1} + \dfrac{4}{(x - 1)^2}$ Check at $x = 2$ the left side is $\tfrac{7}{2}$ and the right side is $\tfrac32 - 2 + 4 = \tfrac72$

Problem 5

Decompose $\dfrac{2x^2 + 3x + 3}{(x + 1)(x^2 + 1)}$

$\dfrac{2x^2 + 3x + 3}{(x + 1)(x^2 + 1)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 1}$ $x^2 + 1$ has no real roots, so it stays as a factor, and its numerator is a general linear polynomial $Bx + C$ not a constant.
$2x^2 + 3x + 3 = A(x^2 + 1) + (Bx + C)(x + 1)$ Multiply both sides by $(x + 1)(x^2 + 1)$
$x = -1$ $2 - 3 + 3 = 2A$ so $A = 1$ The root of the linear factor.
$x^2$ $2 = A + B$ so $B = 1$ Constant: $3 = A + C$ so $C = 2$ Match coefficients. $(Bx + C)(x + 1) = Bx^2 + (B + C)x + C$
$\dfrac{1}{x + 1} + \dfrac{x + 2}{x^2 + 1}$ Check the $x$ coefficient, which was not used: $B + C = 3$ as required.

Problem 6

Write $\dfrac{x^3 + x^2 + x + 3}{x^2 - 1}$ as a polynomial plus partial fractions.

The fraction is improper: degree 3 over degree 2.Divide before decomposing.
$x^3 + x^2 + x + 3 = (x + 1)(x^2 - 1) + (2x + 4)$ Long division: $x \cdot (x^2 - 1) = x^3 - x$ leaves $x^2 + 2x + 3$ $1\cdot(x^2 - 1)$ leaves $2x + 4$ of degree less than 2. Quotient $x + 1$ remainder $2x + 4$
$\dfrac{x^3 + x^2 + x + 3}{x^2 - 1} = x + 1 + \dfrac{2x + 4}{(x - 1)(x + 1)}$ Step 2 divided by $x^2 - 1$
$\dfrac{2x + 4}{(x - 1)(x + 1)} = \dfrac{3}{x - 1} - \dfrac{1}{x + 1}$ Cover-up: at $x = 1$ $\tfrac{6}{2} = 3$ at $x = -1$ $\tfrac{2}{-2} = -1$
$x + 1 + \dfrac{3}{x - 1} - \dfrac{1}{x + 1}$ Check at $x = 2$ the left side is $\tfrac{17}{3}$ and the right side is $3 + 3 - \tfrac13 = \tfrac{17}{3}$

Problem 7

Find $\displaystyle\int\frac{5x - 1}{(x + 1)(x - 2)}\,dx$

$\displaystyle\int\left(\frac{2}{x + 1} + \frac{3}{x - 2}\right)dx$ The decomposition from Problem 1.
$\displaystyle\int\frac{2}{x + 1}dx = 2\ln|x + 1|$ and $\displaystyle\int\frac{3}{x - 2}dx = 3\ln|x - 2|$ $\int\tfrac{dx}{x - a} = \ln|x - a|$ with the constant carried along. The absolute value keeps the log defined on both sides of the root.
$2\ln|x + 1| + 3\ln|x - 2| + C$ Check by differentiating: $\tfrac{2}{x + 1} + \tfrac{3}{x - 2}$ which is the integrand.

Problem 8

Evaluate $\displaystyle\int_2^3\frac{dx}{x^2 - 1}$ exactly.

$\dfrac{1}{(x - 1)(x + 1)} = \dfrac{1/2}{x - 1} - \dfrac{1/2}{x + 1}$ Cover-up: at $x = 1$ $\tfrac{1}{2}$ at $x = -1$ $\tfrac{1}{-2}$
$\displaystyle\int\frac{dx}{x^2 - 1} = \tfrac12\big(\ln|x - 1| - \ln|x + 1|\big) = \tfrac12\ln\left|\frac{x - 1}{x + 1}\right|$ Integrate each term, then combine with $\ln p - \ln q = \ln\tfrac pq$
At $x = 3$ $\tfrac12\ln\tfrac24 = \tfrac12\ln\tfrac12$ At $x = 2$ $\tfrac12\ln\tfrac13$ Neither root of $x^2 - 1$ lies in $[2, 3]$ so the integrand is continuous there and the antiderivative can be evaluated at the ends.
$\tfrac12\ln\tfrac12 - \tfrac12\ln\tfrac13 = \tfrac12\ln\dfrac{1/2}{1/3}$ Upper value minus lower value, then the same log rule.
$\tfrac12\ln\tfrac32 \approx 0.2027$ Sanity check: the integrand falls from $\tfrac13$ to $\tfrac18$ on an interval of length 1, so the integral lies between $0.125$ and $0.333$

Problem 9

Find $\displaystyle\int\frac{x^2 + 3}{x(x - 1)^2}\,dx$

$\displaystyle\int\left(\frac{3}{x} - \frac{2}{x - 1} + \frac{4}{(x - 1)^2}\right)dx$ The decomposition from Problem 4.
$\displaystyle\int\frac{3}{x}dx = 3\ln|x|$ and $\displaystyle\int\frac{-2}{x - 1}dx = -2\ln|x - 1|$ The two linear terms give logs.
$\displaystyle\int\frac{4}{(x - 1)^2}dx = 4\int(x - 1)^{-2}dx = -\frac{4}{x - 1}$ The squared term is a power, not a log: $\int u^{-2}du = -u^{-1}$ with $u = x - 1$
$3\ln|x| - 2\ln|x - 1| - \dfrac{4}{x - 1} + C$ Check the last term: $\tfrac{d}{dx}\big(-4(x - 1)^{-1}\big) = 4(x - 1)^{-2}$

Problem 10

Find $\displaystyle\int\frac{2x^2 + 3x + 3}{(x + 1)(x^2 + 1)}\,dx$

$\displaystyle\int\left(\frac{1}{x + 1} + \frac{x}{x^2 + 1} + \frac{2}{x^2 + 1}\right)dx$ The decomposition from Problem 5, with $\tfrac{x + 2}{x^2 + 1}$ split into its $x$ part and its constant part, which integrate differently.
$\displaystyle\int\frac{x}{x^2 + 1}dx = \tfrac12\ln(x^2 + 1)$ Substitution $u = x^2 + 1$ $du = 2x\,dx$ the numerator is half of $du$ No absolute value is needed because $x^2 + 1 > 0$
$\displaystyle\int\frac{2}{x^2 + 1}dx = 2\arctan x$ The constant part gives the arctangent.
$\ln|x + 1| + \tfrac12\ln(x^2 + 1) + 2\arctan x + C$ Steps 2 and 3 plus $\int\tfrac{dx}{x + 1} = \ln|x + 1|$

Where this goes wrong

1. Decomposing an improper fraction without dividing

The cover-up rule produces numbers for any fraction, proper or not, so nothing seems to go wrong.

$\dfrac{x^3 + x^2 + x + 3}{(x - 1)(x + 1)}$ Right so far: Problem 6 with the denominator factored.
“Two linear factors, so $\tfrac{A}{x - 1} + \tfrac{B}{x + 1}$ cover up to get $A = \tfrac62 = 3$ and $B = \tfrac{2}{-2} = -1$ ”The habit that causes the mistake: going straight to the form without comparing degrees.
$\dfrac{x^3 + x^2 + x + 3}{x^2 - 1} = \dfrac{3}{x - 1} - \dfrac{1}{x + 1}$ The right side goes to 0 as $x \to \infty$ while the left side grows like $x$ so they cannot be equal; at $x = 2$ they are $\tfrac{17}{3}$ and $\tfrac83$ A proper-fraction form can only match a proper fraction. Divide first; the missing piece is the quotient $x + 1$ (Problem 6).

2. One term for a repeated factor

$(x - 1)^2$ looks like a single factor, so it gets a single fraction.

$\dfrac{x^2 + 3}{x(x - 1)^2}$ Right so far: Problem 4.
“One fraction per factor: $x$ and $(x - 1)^2$ ”The habit that causes the mistake: counting factors, not powers.
$\dfrac{x^2 + 3}{x(x - 1)^2} = \dfrac{A}{x} + \dfrac{B}{(x - 1)^2}$ Multiplying out gives $x^2 + 3 = A(x - 1)^2 + Bx$ The roots give $A = 3$ and $B = 4$ but then the $x^2$ coefficients say $1 = A = 3$ no constants work. A repeated factor needs one term for each power, $\tfrac{B}{x - 1} + \tfrac{C}{(x - 1)^2}$ (Problem 4).

3. A constant over an irreducible quadratic

Every linear factor gets a constant on top, and that pattern gets carried to $x^2 + 1$

$\dfrac{2x^2 + 3x + 3}{(x + 1)(x^2 + 1)}$ Right so far: Problem 5.
“Put a constant over each factor.”The habit that causes the mistake: the numerator over a quadratic needs one degree less than the quadratic, which is degree 1, not 0.
$\dfrac{2x^2 + 3x + 3}{(x + 1)(x^2 + 1)} = \dfrac{A}{x + 1} + \dfrac{C}{x^2 + 1}$ Matching coefficients of $2x^2 + 3x + 3 = A(x^2 + 1) + C(x + 1)$ gives $A = 2$ from $x^2$ $C = 3$ from $x$ and then $3 = A + C = 5$ from the constants: no solution. With $\tfrac{Bx + C}{x^2 + 1}$ there are three unknowns for three equations (Problem 5).

4. Evaluating the cover-up at the wrong root

The factor is $x + 2$ and 2 is the number in it.

$\dfrac{x + 7}{(x - 3)(x + 2)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 2}$ with $A = 2$ Right so far: Problem 2, steps 2 and 3.
“Cover up $x + 2$ and put in $x = 2$ ”The habit that causes the mistake: reading the number off the factor without its sign.
$B = \dfrac{2 + 7}{2 - 3} = -9$ The cover-up works because the covered factor is 0 at the chosen $x$ and $x + 2$ is 0 at $x = -2$ not 2. At $x = -2$ $B = \tfrac{5}{-5} = -1$ A quick check catches it: $2(x + 2) - 9(x - 3) = -7x + 31$ not $x + 7$

5. Integrating 1/(x − 1)² as a log

Every other term in the decomposition integrates to a log, so this one does too.

$\displaystyle\int\left(\frac{3}{x} - \frac{2}{x - 1} + \frac{4}{(x - 1)^2}\right)dx$ Right so far: Problem 9, step 1.
“ $\int\tfrac{du}{u} = \ln|u|$ with $u = (x - 1)^2$ ”The habit that causes the mistake: $\int\tfrac{du}{u}$ needs $du$ on top, and $du = 2(x - 1)\,dx$ is not there.
$\displaystyle\int\frac{4}{(x - 1)^2}dx = 4\ln(x - 1)^2$ Differentiating the claimed answer gives $\tfrac{8}{x - 1}$ not $\tfrac{4}{(x - 1)^2}$ The term is the power $4(x - 1)^{-2}$ whose integral is $-\tfrac{4}{x - 1}$ (Problem 9).

Print this set: partial-fractions.pdf (problems, answers, and worked solutions on separate pages).

Read the guide: Partial fraction decomposition: pick the right form, then find A and B