Integral Calculus 201-SN3-RE • CEGEP, Montreal

Revision sheet: partial fraction decomposition (201-SN3-RE)

Q: When do you need to divide before decomposing a rational function?

As soon as the degree of the numerator is greater than or equal to that of the denominator. A partial fraction decomposition only exists for a fraction whose numerator has a strictly smaller degree. So you carry out the polynomial long division, keep the quotient as it is in the final answer, and decompose only the remainder over the original denominator.

Q: How do you know how many terms to write for a squared factor?

A linear factor raised to the power m opens exactly m terms, one for each power from 1 to m. A squared factor therefore gives two terms, one over the factor alone and one over the factor squared. The check is immediate: the total number of unknowns must equal the degree of the factored denominator.

Q: When do you use a numerator of the form Bx plus C?

Only over an irreducible quadratic factor, that is, one whose discriminant is strictly negative. If the discriminant is positive or zero, the quadratic factors into linear factors and each of those factors then gets a constant numerator. So the discriminant is computed before writing a single term.

Q: How does the cover-up method work and when is it not allowed?

For a simple pole a, cover the factor x minus a in the denominator and evaluate everything else at a: the result is directly the coefficient of that term. It also works on the highest power of a repeated factor. It is not allowed on an intermediate power or on an irreducible factor, where a zero factor would remain in the denominator.

Q: How do you check a partial fraction decomposition?

Choose a simple integer value that does not make the denominator vanish, then compute the original fraction and the decomposition at that value. Since the decomposition is unique, a single well-chosen value is enough to detect almost every sign or coefficient error, without redoing any computation.

Q: Why does a fraction with a factor x squared plus 1 give an arctangent?

Because the numerator of degree 1 splits into two parts. The part in x has the form derivative over function and gives a logarithm. The constant part is, up to a coefficient, the derivative of the arctangent. A single term therefore produces two antiderivatives of different kinds, and forgetting the second one is the most frequent mistake of the chapter.

Partial fraction decomposition is not a computing technique, it is a READING technique: you look at the factored denominator and write the form, empty terms included, before touching a single coefficient. That is why the papers that are lost are lost on the first line, and why rereading them does not help.

The chapter exists for one reason only in Integral Calculus: finding the antiderivative of a rational function. Each term of the decomposition matches a known antiderivative, $\frac{A}{x-a}$ a logarithm, $\frac{A}{(x-a)^m}$ a power, $\frac{Bx+C}{x^2+1}$ a logarithm PLUS an arctangent. That last line is the one that costs the most marks in exams.

This sheet is written for CEGEP students in the Science program (Sciences de la nature) taking Integral Calculus, 201-SN3-RE (formerly 201-NYB). Once it is read, the corrected exercise set on the same chapter puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The whole chapter is decided BEFORE the first coefficient: the form is read from the factored denominator, and a wrong form makes everything that follows wrong, however carefully it is computed.

This chapter is part of 201-SN3-RE, Integral Calculus

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Rational function
Polynomial long division
Partial fraction decomposition
Simple pole
Repeated linear factor
Irreducible quadratic factor
Discriminant
Cover-up method
Equating coefficients
Uniqueness of the decomposition
Antiderivative of a rational function

The essentials

The required order: four moves before the first coefficient

• DIVIDE FIRST. If $\deg(A)\geq\deg(B)$ , no decomposition exists in the expected form. Divide, get $\dfrac{A}{B}=Q+\dfrac{R}{B}$ with $\deg(R)<\deg(B)$ , and decompose $\dfrac{R}{B}$ .
• FACTOR THE DENOMINATOR COMPLETELY. A factor left as $x^2-1$ or $x^4-1$ is not factored: it still splits into $(x-1)(x+1)$ and $(x-1)(x+1)(x^2+1)$ .
• OPEN THE TERMS. Each factor of the denominator opens a number of terms fixed by its kind and its power, and those terms are written BEFORE any computation.
• COUNT. The total number of unknowns must equal the degree of the denominator. It is the fastest check that a form is correct, and it takes two seconds.

The first three moves produce no number: they produce the FORM. Computing the coefficients only comes afterwards, and it never rescues a wrong form.

The decomposition exists and it is UNIQUE. Two different forms of the same result are therefore the same up to rewriting, and one well-chosen numerical value is enough to settle it.

One factor of the denominator, one pole, one term

• Each factor $(x-a)$ of the denominator gives a pole at $a$ (a vertical asymptote on the graph) and a term of the decomposition. The terms do not mix: they add up.
• $\dfrac{1}{x^2-1}=\dfrac{1}{(x-1)(x+1)}=\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}$ , not $\dfrac{1}{x-1}-\dfrac{1}{x+1}$ : the coefficients are not $1$ by default.
• An irreducible quadratic factor gives no real pole, but it still gives its term, with a numerator of degree $1$ : $\dfrac{Bx+C}{ax^2+bx+c}$ .

Two factors in the denominator, two vertical asymptotes:

x-1

gives the pole at

1

and the term

\dfrac{1/2}{x-1}

x+1

gives the pole at

-1

and the term

-\dfrac{1/2}{x+1}

What each term is for, once the decomposition is written

• $\displaystyle\int\dfrac{A}{x-a}\,dx=A\ln|x-a|+K$ . The absolute value is not optional.
• $\displaystyle\int\dfrac{A}{(x-a)^m}\,dx=\dfrac{A}{(1-m)(x-a)^{m-1}}+K$ for $m\geq 2$ : this is NOT a logarithm.
• $\displaystyle\int\dfrac{Bx+C}{x^2+1}\,dx=\dfrac{B}{2}\ln(x^2+1)+C\arctan(x)+K$ : two antiderivatives of different kinds in a single term.

That is the only reason the chapter exists in integral calculus: a decomposed rational function is a sum of known antiderivatives, and nothing else.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Decomposing without dividing when the numerator has a degree at least as large

the whole question

What not to write

“ $\dfrac{x^4+3x^2+2x+4}{x^3-x^2}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-1}$ .”

What to write

“ $\deg(A)=4\geq\deg(B)=3$ , so I divide first: $\dfrac{x^4+3x^2+2x+4}{x^3-x^2}=x+1+\dfrac{4x^2+2x+4}{x^3-x^2}$ , and it is the remainder that I decompose.”

The division stops when the degree of the remainder falls below that of the divisor:

4x^2+2x+4

is left, of degree

2

, and the quotient

x+1

is copied as is into the final answer.

Why: The resulting system is inconsistent, so the student goes round in circles for ten minutes, then invents a coefficient. Comparing the two degrees takes one second and comes BEFORE everything.

2. Setting up a numerator $Bx+C$ without looking at the discriminant

the whole question

What not to write

“ $\dfrac{5}{x^2+x-6}=\dfrac{Bx+C}{x^2+x-6}$ , it is a quadratic so the numerator has degree $1$ .”

What to write

“ $\Delta=1+24=25>0$ , so $x^2+x-6=(x-2)(x+3)$ and the form is $\dfrac{A}{x-2}+\dfrac{B}{x+3}$ , with constant numerators.”

The same quadratic up to one sign: on the left

\Delta>0

opens two terms with constant numerators, on the right

\Delta<0

opens only one, with a numerator of degree

1

Why: A quadratic keeps its numerator of degree $1$ only if it is IRREDUCIBLE. The discriminant is the only thing that tells you, and it is computed in your head.

3. Opening only one term for a repeated factor

3 marks, and a system with no solution

What not to write

“ $\dfrac{3x+1}{x^2(x-1)}=\dfrac{A}{x^2}+\dfrac{B}{x-1}$ .”

What to write

“ $x^2$ is a repeated factor with exponent $2$ , it opens TWO terms: $\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-1}$ , that is, three unknowns for a denominator of degree $3$ .”

0

, a DOUBLE pole: the curve plunges on the same side to the left and to the right. At

1

, a simple pole: it changes sign. Two behaviours, so two terms for

x^2

Why: The number of unknowns must equal the degree of the denominator. With two unknowns for degree $3$ , the form is mathematically impossible, however carefully the computation is done.

4. Applying the cover-up method to an intermediate power

2 marks, and a division by zero written on the paper

What not to write

“For $\dfrac{A}{x}$ in $\dfrac{3x+1}{x^2(x-1)}$ , I cover $x$ and evaluate at $0$ : $A=\dfrac{3\cdot 0+1}{0\cdot(0-1)}$ .”

What to write

“The cover-up method reaches $\dfrac{B}{x^2}$ , the HIGHEST power, by covering $x^2$ and evaluating at $0$ : $B=\dfrac{1}{-1}=-1$ . For $A$ , I equate coefficients or give $x$ a simple value.”

Why: The cover-up method relies on a finite limit at $a$ : it exists only for a simple pole or for the highest power of a repeated factor. Everywhere else an $x-a$ is left in the denominator.

5. Believing the coefficients are $1$ when the numerator is $1$

1 mark, then an antiderivative off by a factor of

2

What not to write

“ $\dfrac{1}{x^2-1}=\dfrac{1}{x-1}-\dfrac{1}{x+1}$ .”

What to write

“ $\dfrac{1}{x^2-1}=\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}$ : cover-up at $1$ gives $\dfrac{1}{1+1}=\dfrac12$ , and cover-up at $-1$ gives $\dfrac{1}{-1-1}=-\dfrac12$ .”

In green,

\frac{1}{x^{2}-1}

; in red,

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

, which equals

\frac{2}{x^{2}-1}

: twice too high everywhere. The coefficients are

\pm\frac{1}{2}

Why: The mistake survives every rereading because the form is right and only the numbers are wrong. An evaluation at $x=3$ catches it in five seconds.

6. Putting a constant numerator over an irreducible quadratic factor

the whole question

What not to write

“ $\dfrac{2x+5}{(x-1)(x^2+4)}=\dfrac{A}{x-1}+\dfrac{B}{x^2+4}$ .”

What to write

“ $x^2+4$ is irreducible and of degree $2$ , so it carries a numerator of degree $1$ : $\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+4}$ , that is, three unknowns for a denominator of degree $3$ .”

Why: The numerator of a term always has a degree strictly lower than that of its factor. A factor of degree $2$ therefore calls for $Bx+C$ , and a factor of degree $1$ for a constant.

7. Stopping at a half-done factorization

the whole question

What not to write

“ $\dfrac{1}{x^4-1}=\dfrac{A}{x^2-1}+\dfrac{B}{x^2+1}$ .”

What to write

“ $x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)$ , so the form is $\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{Cx+D}{x^2+1}$ : four unknowns for degree $4$ .”

Why: The count gives the omission away at once: the first form has only two unknowns for a denominator of degree $4$ . Any difference of squares factors further.

8. Integrating $\dfrac{Bx+C}{x^2+1}$ as a single logarithm

2 marks, every time

What not to write

“ $\displaystyle\int\dfrac{3x+2}{x^2+1}\,dx=\dfrac32\ln(x^2+1)+K$ .”

What to write

“I split: $\displaystyle\int\dfrac{3x}{x^2+1}\,dx+\int\dfrac{2}{x^2+1}\,dx=\dfrac32\ln(x^2+1)+2\arctan(x)+K$ .”

Why: Only the part in $Bx$ has the form $\dfrac{u'}{u}$ . The constant $C$ fits into no logarithm and always produces an arctangent; it is the most frequent mistake of the chapter in exams.

9. Handing in the paper without evaluating the decomposition at one point

2 to 5 marks, depending on the number of terms

What not to write

“I found $A$ , $B$ and $C$ , the system was right, I move on to the next exercise.”

What to write

“I check at $x=2$ : the original fraction equals $9$ and my decomposition $2+1-3-1+10=9$ . I can hand it in.”

Why: Uniqueness of the decomposition makes the test decisive: one well-chosen value, away from the poles, catches almost every sign and coefficient error.

Which method to choose

How many terms each factor of the denominator opens

Look ONLY at the shape of the factor, once the denominator is completely factored.

The dictionary of the chapter: on the left what you see in the denominator, in the middle what you write, on the right the number of unknowns, whose total must equal the degree of the denominator.

If simple linear factor $(x-a)$ → one term $\dfrac{A}{x-a}$ , one unknown

Example: $\dfrac{1}{(x-1)(x+2)}\to\dfrac{A}{x-1}+\dfrac{B}{x+2}$
If repeated linear factor $(x-a)^m$ → $m$ terms, from $\dfrac{A_1}{x-a}$ to $\dfrac{A_m}{(x-a)^m}$ , so $m$ unknowns

Example: $x^2(x-1)\to\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-1}$

the term of lowest power is the one that gets forgotten, never the one of highest power
If quadratic $ax^2+bx+c$ with $\Delta<0$ → one term $\dfrac{Bx+C}{ax^2+bx+c}$ , two unknowns

Example: $\dfrac{1}{(x-1)(x^2+4)}\to\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+4}$
If repeated irreducible quadratic $(ax^2+bx+c)^m$ → $m$ terms with numerators of degree $1$ , so $2m$ unknowns

Example: $(x^2+1)^2\to\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}$
If quadratic with $\Delta\geq 0$ → not a separate case: factor it and you are back to linear factors

Example: $x^2+x-6=(x-2)(x+3)$

No branch applies as long as $\deg(A)\geq\deg(B)$ : in that case you divide, and apply the tree to the remainder. Final check: the total number of unknowns equals the degree of the denominator.

How to get each coefficient without solving a full system

Look at the term whose coefficient you want, not at the whole fraction.

If term of a simple pole $\dfrac{A}{x-a}$ → cover-up: hide $(x-a)$ in the denominator and evaluate what is left at $a$

Example: in $\dfrac{1}{(x-1)(x+2)}$ , $A=\dfrac{1}{1+2}=\dfrac13$
If term of the highest power of a repeated factor → cover-up too: hide the whole $(x-a)^m$ and evaluate at $a$

Example: in $\dfrac{3x+1}{x^2(x-1)}$ , $B=\dfrac{1}{-1}=-1$
If term of an intermediate power → no cover-up: equate one coefficient of the same degree, or give $x$ a simple value

Example: the coefficient of $x^2$ gives $A$ in one line
If term of an irreducible factor → equate coefficients: expand and match the coefficients of each degree

the extreme degrees, the constant term and the leading term, give the shortest equations
If only one or two unknowns are left → convenient values: $x=0$ , $x=1$ , $x=-1$ , away from the poles

Example: $x=0$ isolates the constant term in one operation

The cover-up method is a shortcut, never a complete method: on a decomposition with four terms, it typically gives two, and the other two come from equating coefficients.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Decomposing a rational function, from the first mark to the last

When to use it: The question says “decompose into partial fractions”, or asks for an antiderivative whose denominator factors.

1 Compare the degrees and write the conclusion: “ $\deg(A)<\deg(B)$ , the decomposition exists as it stands”, or carry out the division and start again from the remainder.
2 Factor the denominator completely and justify the irreducibility of each quadratic you keep with its discriminant.
3 Set up the form, every term written out, then count: “ $3$ unknowns for a denominator of degree $3$ ”.
4 Bring to a common denominator and write the equality of the numerators, valid for every $x$ except at the poles.
5 Get the coefficients the cover-up method can reach, then the others by equating coefficients or by convenient values.
6 Check at a simple point away from the poles, and write it on the paper.

Concluding sentence

“For every $x\notin\{0, 1\}$ , $\dfrac{x^4+3x^2+2x+4}{x^3-x^2}=x+1-\dfrac{6}{x}-\dfrac{4}{x^2}+\dfrac{10}{x-1}$ .”

The trap: Leaving the quotient of the division out of the final answer. It is part of the equality, not scratch work.

Marking: Typically $1$ mark for the complete form, $1$ mark per coefficient, $1$ mark for the conclusion with its domain.

Moving on to the antiderivative once the decomposition is written

When to use it: The question asks for $\displaystyle\int\dfrac{A(x)}{B(x)}\,dx$ and the denominator factors.

1 Decompose first, completely, and state that the integral is computed term by term by linearity.
2 Integrate each term with its own rule: logarithm with absolute value for a simple pole, power rule for a higher power, logarithm PLUS arctangent for an irreducible factor.
3 Combine the logarithms into one if the question asks for it, never otherwise.
4 Add the constant and, for a definite integral, check that no pole lies in the interval before evaluating at the bounds.

Concluding sentence

“ $\displaystyle\int\dfrac{3x+2}{x^2+1}\,dx=\dfrac32\ln(x^2+1)+2\arctan(x)+K$ .”

The trap: Evaluating a definite integral whose interval contains a pole. The integral is then improper and the direct computation gives a wrong number.

Marking: $1$ mark per antiderivative of a term, $1$ mark for the absolute value, $1$ mark for the arctangent.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Count the unknowns against the degree of the denominator

Before any computation, add up the unknowns you set up and compare with the degree of the factored denominator. The two numbers must be equal.

$\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-1}$ sets up $3$ unknowns for $x^2(x-1)$ , of degree $3$ : the form is possible. Two unknowns would have been enough to declare it wrong.
Evaluate at a simple point, away from the poles

Choose an integer $x$ that does not make the denominator vanish, compute the original fraction and the decomposition, and compare. One value is almost always enough.

At $x=2$ : the fraction equals $\dfrac{36}{4}=9$ and the decomposition $2+1-3-1+10=9$ . A gap of $2$ points to a coefficient, a sign gap to a cover-up error.
Compare the behaviour at infinity

As $x\to\infty$ , each term tends to $0$ and only the quotient of the division is left. The original fraction must behave like that quotient.

$\dfrac{x^4+3x^2+2x+4}{x^3-x^2}$ behaves like $x$ , and the decomposition leaves $x+1$ : consistent. A decomposition without a quotient would have tended to $0$ , which is visibly wrong.
Differentiate the antiderivative you found

On an integration question, differentiate the result in your head and check that you get back the integrand. It is faster than redoing the computation.

Differentiating $\dfrac32\ln(x^2+1)$ gives $\dfrac{3x}{x^2+1}$ , not $\dfrac{3x+2}{x^2+1}$ : the arctangent is missing, and the mistake shows without recomputing anything.

The typical problem, taken apart

A fraction that gathers the three difficulties of the chapter

Decompose $\dfrac{x^4+3x^2+2x+4}{x^3-x^2}$ into partial fractions, then check the result.

This fraction combines the three difficulties: a numerator of higher degree, a denominator to factor, and a repeated factor.

Step 1

$\deg(A)=4$ and $\deg(B)=3$ , so $\deg(A)\geq\deg(B)$ : the decomposition does not exist in the expected form. Long division: $x^4+3x^2+2x+4=(x+1)(x^3-x^2)+4x^2+2x+4$ , so $\dfrac{x^4+3x^2+2x+4}{x^3-x^2}=x+1+\dfrac{4x^2+2x+4}{x^3-x^2}$ .

Why

Comparing the degrees is the very first move, even before looking at the denominator. Without it, the system set up below would be inconsistent and the exercise would be lost without a single wrong computation.

Step 2

$x^3-x^2=x^2(x-1)$ . The factor $x$ is repeated with exponent $2$ , the factor $x-1$ is simple. The form is therefore $\dfrac{4x^2+2x+4}{x^2(x-1)}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-1}$ , that is, $3$ unknowns for a denominator of degree $3$ .

Why

Write every term before looking for a single coefficient, and state the count. This is the line the marker looks for first, and it is the one worth the most marks for the work it takes.

Step 3

Over a common denominator: $4x^2+2x+4=Ax(x-1)+B(x-1)+Cx^2$ , an equality valid for every $x\neq 0$ and $x\neq 1$ , hence for every $x$ by continuity of polynomials.

Why

The equality of the numerators is what allows you to substitute $x=0$ and $x=1$ , even though they are excluded from the original domain. Saying so in one sentence avoids the marker's comment.

Step 4

$x=0$ : $4=-B$ , so $B=-4$ . $x=1$ : $4+2+4=C$ , so $C=10$ . Coefficient of $x^2$ : $4=A+C$ , so $A=4-10=-6$ .

Why

The two convenient values give two coefficients without any system, and matching the leading coefficient gives the third in one line. Solving a $3\times 3$ system here would be a pure waste of time.

Step 5

Check at $x=2$ : the original fraction equals $\dfrac{16+12+4+4}{8-4}=\dfrac{36}{4}=9$ , and the decomposition equals $2+1-\dfrac{6}{2}-\dfrac{4}{4}+\dfrac{10}{1}=3-3-1+10=9$ .

Why

A single evaluation covers all four terms at once, quotient included. It is the most profitable check of the chapter, and it fits on one line of the paper.

Step 6

Antiderivative, by linearity: $\displaystyle\int\dfrac{x^4+3x^2+2x+4}{x^3-x^2}\,dx=\dfrac{x^2}{2}+x-6\ln|x|+\dfrac{4}{x}+10\ln|x-1|+K$ .

Why

Each term gives its own antiderivative: $\dfrac{-4}{x^2}$ gives $\dfrac{4}{x}$ and certainly NOT a logarithm, and both logarithms keep their absolute value. This is where the decomposition finally pays off.

The conclusion, written out

“For every $x\notin\{0, 1\}$ , $\dfrac{x^4+3x^2+2x+4}{x^3-x^2}=x+1-\dfrac{6}{x}-\dfrac{4}{x^2}+\dfrac{10}{x-1}$ , a result checked at $x=2$ .”

The classic mistake on this problem: The classic mistake on this exercise is writing the decomposition without the quotient $x+1$ , either because the division was forgotten, or because it was done as scratch work and never copied out. The check at $x=2$ detects it at once: without the quotient, you get $6$ instead of $9$ .

Learn by heart

• If $\deg(A)\geq\deg(B)$ , DIVIDE before decomposing, and the quotient is part of the answer.
• A quadratic keeps its numerator $Bx+C$ only if $\Delta<0$ . Otherwise, factor it.
• $(x-a)^m$ opens $m$ terms, from $\dfrac{A_1}{x-a}$ to $\dfrac{A_m}{(x-a)^m}$ .
• The total number of unknowns equals the DEGREE of the denominator, always.
• The cover-up method reaches a simple pole and the highest power of a repeated factor, never an intermediate power nor an irreducible factor.
• $\displaystyle\int\dfrac{Bx+C}{x^2+1}\,dx=\dfrac{B}{2}\ln(x^2+1)+C\arctan(x)+K$ : a logarithm AND an arctangent.
• The decomposition is UNIQUE, so an evaluation at one point away from the poles is enough to validate it.

Frequently asked questions

When do you need to divide before decomposing a rational function?

As soon as the degree of the numerator is greater than or equal to that of the denominator. A partial fraction decomposition only exists for a fraction whose numerator has a strictly smaller degree. So you carry out the polynomial long division, keep the quotient as it is in the final answer, and decompose only the remainder over the original denominator.

How do you know how many terms to write for a squared factor?

A linear factor raised to the power m opens exactly m terms, one for each power from 1 to m. A squared factor therefore gives two terms, one over the factor alone and one over the factor squared. The check is immediate: the total number of unknowns must equal the degree of the factored denominator.

When do you use a numerator of the form Bx plus C?

Only over an irreducible quadratic factor, that is, one whose discriminant is strictly negative. If the discriminant is positive or zero, the quadratic factors into linear factors and each of those factors then gets a constant numerator. So the discriminant is computed before writing a single term.

How does the cover-up method work and when is it not allowed?

For a simple pole a, cover the factor x minus a in the denominator and evaluate everything else at a: the result is directly the coefficient of that term. It also works on the highest power of a repeated factor. It is not allowed on an intermediate power or on an irreducible factor, where a zero factor would remain in the denominator.

How do you check a partial fraction decomposition?

Choose a simple integer value that does not make the denominator vanish, then compute the original fraction and the decomposition at that value. Since the decomposition is unique, a single well-chosen value is enough to detect almost every sign or coefficient error, without redoing any computation.

Why does a fraction with a factor x squared plus 1 give an arctangent?

Because the numerator of degree 1 splits into two parts. The part in x has the form derivative over function and gives a logarithm. The constant part is, up to a coefficient, the derivative of the arctangent. A single term therefore produces two antiderivatives of different kinds, and forgetting the second one is the most frequent mistake of the chapter.

Practise it

Corrected exercises: Partial fraction decomposition, 201-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

16 corrected exercises
160 points
240 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Revision sheet: partial fraction decomposition (201-SN3-RE)

The essentials

The required order: four moves before the first coefficient

One factor of the denominator, one pole, one term

What each term is for, once the decomposition is written

The mistakes that cost marks

1. Decomposing without dividing when the numerator has a degree at least as large

2. Setting up a numerator Bx+CBx+CBx+C without looking at the discriminant

3. Opening only one term for a repeated factor

4. Applying the cover-up method to an intermediate power

5. Believing the coefficients are 111 when the numerator is 111

6. Putting a constant numerator over an irreducible quadratic factor

7. Stopping at a half-done factorization

8. Integrating Bx+Cx2+1\dfrac{Bx+C}{x^2+1}x2+1Bx+C​ as a single logarithm

9. Handing in the paper without evaluating the decomposition at one point

Which method to choose

How many terms each factor of the denominator opens

How to get each coefficient without solving a full system

How the answer is expected to be written

Decomposing a rational function, from the first mark to the last

Moving on to the antiderivative once the decomposition is written

Check before you hand in

The typical problem, taken apart

A fraction that gathers the three difficulties of the chapter

Learn by heart

Frequently asked questions

When do you need to divide before decomposing a rational function?

How do you know how many terms to write for a squared factor?

When do you use a numerator of the form Bx plus C?

How does the cover-up method work and when is it not allowed?

How do you check a partial fraction decomposition?

Why does a fraction with a factor x squared plus 1 give an arctangent?

Corrected exercises: Partial fraction decomposition, 201-SN3-RE

See also

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2. Setting up a numerator $Bx+C$ without looking at the discriminant

5. Believing the coefficients are $1$ when the numerator is $1$

8. Integrating $\dfrac{Bx+C}{x^2+1}$ as a single logarithm