MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: partial fractions (MATH 141)

Q: When do I need long division before partial fractions?

Whenever the degree of the numerator is greater than or equal to the degree of the denominator. Partial fractions only decompose a proper fraction. Divide first, keep the polynomial quotient, which integrates directly, and decompose only the remainder over the denominator. Skipping the division loses the quotient, even when the cover-up seems to give sensible constants.

Q: How do I know if a quadratic factor is irreducible?

Compute its discriminant, b squared minus four a c. If it is negative, the quadratic has no real root and stays whole in the decomposition, with a linear numerator on top, and it will give an arctangent. If it is positive, it factors into two linear factors and gives logarithms. If it is zero, it is a perfect square, a repeated factor that gives a power.

Q: Why does a repeated factor need one term for each power?

Because the number of unknown constants must equal the degree of the denominator for the system to have a solution. A squared factor has degree two, so it needs two terms, one over the factor and one over its square. With only the squared term, the fraction cannot blow up at the right rate and the equations contradict each other.

Q: What is the cover-up method and when does it fail?

Multiply the decomposition by one linear factor and evaluate at its root: every other term vanishes and the constant appears. It works for each simple linear factor and for the highest power of a repeated one. It cannot find the lower powers, the numerators over irreducible quadratics, or the missing quotient of an improper fraction; use coefficients or extra values for those.

Q: Why is there an arctangent in some partial fraction integrals?

An irreducible quadratic becomes, after completing the square, a sum of a square and a positive constant. A constant over that sum is the derivative of an arctangent: the integral of one over u squared plus a squared is one over a times the arctangent of u over a. The part of the numerator proportional to the derivative of the quadratic gives a logarithm instead.

This sheet is not a summary of section 7.4 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on partial fractions in MATH 141 at McGill University, and which precise gesture avoids each loss.

Everything below is done by hand, as on the exam, where no calculator is allowed: every decomposition quoted has been checked at an extra value of $x$ , and every antiderivative by differentiating it.

The thread of the chapter

The SHAPE of the answer is decided before any constant, in a fixed order: degree check (divide when $\deg N \ge \deg D$ ), complete factorization (the discriminant decides every quadratic), then one term per power, with a linear numerator over each irreducible quadratic. Each kind of term has one kind of antiderivative: a logarithm, a power, or a logarithm plus an arctangent.

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

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

Topics covered

Partial fraction decomposition
Proper rational function
Polynomial long division
Distinct linear factors
Repeated linear factors
Irreducible quadratic factor
Discriminant of a quadratic
Completing the square
Cover-up method
Rational root theorem
Rationalizing substitution
Inverse tangent integral

The essentials

The order that decides the form

• Step 1, the degree check: if $\deg N \ge \deg D$ , divide first. $\frac{N}{D} = Q + \frac{R}{D}$ with $\deg R < \deg D$ , and ONLY $\frac{R}{D}$ is decomposed. The quotient $Q$ is the curve's behaviour far away: its asymptote.
• Step 2, factor $D$ COMPLETELY over the reals: grouping, the rational root test, differences of squares and cubes. Test every quadratic with its discriminant $b^2 - 4ac$ .
• Step 3, write the form: one term per POWER of each factor. A linear factor takes a constant numerator, an irreducible quadratic a linear numerator $Bx + C$ .
• Step 4, count: the number of unknowns must equal $\deg D$ . Only then look for the constants.
• Step 5, integrate each term with its own rule, then check by differentiating or at one value.

f(x) = \frac{x^2 + 1}{x - 1} = x + 1 + \frac{2}{x - 1}

: the blue curve follows its quotient, the dashed line

x + 1

, while the proper part

\frac{2}{x - 1}

in green dies at infinity.

Most lost marks in this chapter are a step done out of order: constants computed on an improper fraction, a form written on a denominator that still factors, an arctangent over a quadratic nobody tested.

Three kinds of term, three kinds of antiderivative

• Simple linear factor: $\int \frac{A}{px + q}\,dx = \frac{A}{p}\ln|px + q| + K$ . The $\frac{1}{p}$ comes from $du = p\,dx$ .
• Repeated linear factor, power $k \ge 2$ : $\int \frac{A}{(x - a)^k}\,dx = -\frac{A}{(k - 1)(x - a)^{k - 1}} + K$ . A POWER, never a logarithm.
• Irreducible quadratic: complete the square, $(x - h)^2 + a^2$ , split the numerator around the derivative of the quadratic, then $\int \frac{u'}{u} = \ln|u|$ and $\int \frac{du}{u^2 + a^2} = \frac{1}{a}\arctan\frac{u}{a}$ .
• Absolute values stay inside every logarithm of a linear factor: on an interval where $x - 1 < 0$ , $\ln(x - 1)$ does not exist and $\ln|x - 1|$ does.
• The letters $A, B, C$ are taken by the decomposition: call the constant of integration $K$ so the answer is not ambiguous.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Each factor, the terms it opens, and what they integrate to

Read a line as: this factor in the denominator contributes these terms, and they integrate to the last column. Every example below has been checked by differentiation. The red lines are rules students invent.

Factor in the denominator	Terms it contributes	Antiderivative
$x - a$	$\frac{A}{x - a}$	$A\ln\|x - a\|$
Example: $\int \frac{3}{x - 2}\,dx = 3\ln\|x - 2\| + K$ .
$px + q$	$\frac{A}{px + q}$	$\frac{A}{p}\ln\|px + q\|$
Example: $\int \frac{4}{2x + 1}\,dx = 2\ln\|2x + 1\| + K$ , not $4\ln\|2x + 1\|$ .
$(x - a)^2$	$\frac{A}{x - a} + \frac{B}{(x - a)^2}$	$A\ln\|x - a\| - \frac{B}{x - a}$
Example: $\frac{x}{(x - 1)^2} = \frac{1}{x - 1} + \frac{1}{(x - 1)^2}$ integrates to $\ln\|x - 1\| - \frac{1}{x - 1} + K$ .
$x^2 + a^2$	$\frac{Bx + C}{x^2 + a^2}$	$\frac{B}{2}\ln(x^2 + a^2) + \frac{C}{a}\arctan\frac{x}{a}$
Example: $\int \frac{2x + 6}{x^2 + 9}\,dx = \ln(x^2 + 9) + 2\arctan\frac{x}{3} + K$ .
$(x - h)^2 + a^2$	$\frac{Bx + C}{x^2 - 2hx + h^2 + a^2}$	split, then $u = x - h$
Example: $\frac{x}{x^2 - 4x + 5}$ : $x = \frac{1}{2}(2x - 4) + 2$ gives $\frac{1}{2}\ln(x^2 - 4x + 5) + 2\arctan(x - 2) + K$ .
$x^2 + a^2$	$\frac{C}{x^2 + a^2}$	$C\ln(x^2 + a^2)$ no such rule
Example: $\frac{d}{dx}\ln(x^2 + 4) = \frac{2x}{x^2 + 4}$ , not $\frac{1}{x^2 + 4}$ . What to do: A constant numerator over an irreducible quadratic gives $\frac{C}{a}\arctan\frac{x}{a}$ : $\int \frac{dx}{x^2 + 4} = \frac{1}{2}\arctan\frac{x}{2} + K$ .
$(x - a)^2$	$\frac{B}{(x - a)^2}$	$B\ln(x - a)^2$ no such rule
Example: $\frac{d}{dx}\ln(x - 1)^2 = \frac{2}{x - 1}$ , not $\frac{1}{(x - 1)^2}$ . What to do: Use the power rule: $\int (x - a)^{-2}\,dx = -(x - a)^{-1} + K$ .

Every blue line is proved by differentiating its last column. If the derivative does not give back the middle column, the rule does not exist, which is exactly what happens on the two red lines.

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 an improper fraction without dividing first

the term x of the antiderivative, that is the whole length of the interval on a definite integral, and usually 2 to 3 marks

What not to write

“By the cover-up, $\frac{x^2 + 2}{x^2 - 1} = \frac{3/2}{x - 1} - \frac{3/2}{x + 1}$ .”

What to write

“The degrees are equal, so I divide: $\frac{x^2 + 2}{x^2 - 1} = 1 + \frac{3}{x^2 - 1} = 1 + \frac{3/2}{x - 1} - \frac{3/2}{x + 1}$ .”

The true fraction, in blue, tends to

1

; the cover-up answer, dashed in red, tends to

0

. They differ by exactly

1

everywhere: the missing quotient.

Why: The cover-up reads each pole and is blind to the quotient: its constants are right, a whole term is missing. A sum of proper fractions tends to $0$ at infinity, while this fraction tends to $1$ , as the figure shows.

2. Forgetting the inner coefficient of a linear factor

1 mark per term, and every term with a non-monic factor is affected

What not to write

“ $\int \frac{5}{3x - 2}\,dx = 5\ln|3x - 2| + K$ .”

What to write

“With $u = 3x - 2$ , $du = 3\,dx$ : $\int \frac{5}{3x - 2}\,dx = \frac{5}{3}\ln|3x - 2| + K$ .”

Why: Differentiating $5\ln|3x - 2|$ gives $\frac{15}{3x - 2}$ , three times too much. After a cover-up on a factor like $2x - 1$ or $3x + 2$ , the $\frac{1}{p}$ is owed on the integration line.

3. Leaving out the lower power of a repeated factor

the whole question: with a form that is too short, the system has no solution

What not to write

“ $\frac{x + 3}{(x + 1)^2} = \frac{B}{(x + 1)^2}$ , one factor so one term.”

What to write

“ $(x + 1)^2$ opens two boxes: $\frac{x + 3}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2}$ , and $x + 3 = (x + 1) + 2$ gives $A = 1$ , $B = 2$ .”

Why: The number of unknowns must equal the degree of the denominator: here $2$ , and the wrong form has $1$ . A constant $B$ cannot equal the non-constant numerator $x + 3$ . A box that comes out $0$ for some numerator must still be written.

4. Integrating a squared factor into a logarithm

1 to 2 marks, and a wrong value on any definite integral

What not to write

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

What to write

“By the power rule, $\int 2(x + 1)^{-2}\,dx = -\frac{2}{x + 1} + K$ .”

Why: Only the FIRST power gives a logarithm. $\frac{d}{dx}2\ln(x + 1)^2 = \frac{4}{x + 1}$ , which has the wrong power. Differentiating the answer takes five seconds and catches it.

5. Taking a quadratic that factors for an irreducible one

the whole question

What not to write

“ $x^2 + x - 6 = \left(x + \frac{1}{2}\right)^2 - \frac{25}{4}$ , so $\int \frac{dx}{x^2 + x - 6}$ is an arctangent.”

What to write

“The discriminant is $1 + 24 = 25 > 0$ , so $x^2 + x - 6 = (x + 3)(x - 2)$ and $\int \frac{dx}{x^2 + x - 6} = \frac{1}{5}\ln\left|\frac{x - 2}{x + 3}\right| + K$ .”

Why: A completed square with a MINUS sign in front of the constant is a difference of squares: it factors. Only $(x - h)^2 + a^2$ , with a plus, leads to an arctangent. Compute the discriminant before writing any form.

6. Writing a logarithm over an irreducible quadratic

2 marks

What not to write

“ $\int \frac{dx}{x^2 + 6x + 10} = \ln(x^2 + 6x + 10) + K$ .”

What to write

“ $x^2 + 6x + 10 = (x + 3)^2 + 1$ , so $\int \frac{dx}{(x + 3)^2 + 1} = \arctan(x + 3) + K$ .”

Why: A logarithm needs the derivative of the denominator on top, here $2x + 6$ . A constant numerator gives an arctangent after completing the square; a general linear numerator gives both, after splitting it around $2x + 6$ .

7. Stopping the factorization of the denominator too early

1 to 2 marks for the form, then an integral that cannot be finished

What not to write

“ $x^4 - 81 = (x^2 - 9)(x^2 + 9)$ , so $\frac{1}{x^4 - 81} = \frac{Ax + B}{x^2 - 9} + \frac{Cx + D}{x^2 + 9}$ .”

What to write

“ $x^2 - 9$ still splits: $x^4 - 81 = (x - 3)(x + 3)(x^2 + 9)$ , so the form is $\frac{A}{x - 3} + \frac{B}{x + 3} + \frac{Cx + D}{x^2 + 9}$ .”

Why: The form is only valid on a COMPLETE factorization. $\frac{Ax + B}{x^2 - 9}$ is not a final term: it still has to be decomposed, and treating it as an arctangent term would be wrong.

8. Losing du over u after the substitution u = e^x

the whole question: the wrong answer differentiates to the wrong function

What not to write

“With $u = e^x$ : $\int \frac{dx}{e^x + 2} = \int \frac{du}{u + 2} = \ln(e^x + 2) + K$ .”

What to write

“ $du = e^x\,dx$ , so $dx = \frac{du}{u}$ and $\int \frac{du}{u(u + 2)} = \frac{1}{2}\ln\frac{u}{u + 2} = \frac{x}{2} - \frac{1}{2}\ln(e^x + 2) + K$ .”

Why: The substitution converts the WHOLE of $dx$ , not only the integrand. $\frac{d}{dx}\ln(e^x + 2) = \frac{e^x}{e^x + 2}$ : the missing factor $\frac{1}{u}$ is visible at once when the answer is differentiated.

Which method to choose

What to do, by the FORM of the rational function

Look at the degrees, then at each factor of the denominator, before writing any letter

Same

x^2 - 2x

, three constants: the blue parabola crosses the axis twice (two logarithms), the orange one touches it (a power), the green one stays above it (an arctangent).

If $\deg N \ge \deg D$ → long division first, then work on the remainder only

Example: $\frac{x^3}{x^2 - 4} = x + \frac{4x}{x^2 - 4}$
If the numerator is a multiple of the derivative of the denominator → substitution u = denominator, no decomposition

Example: $\int \frac{4x}{x^2 - 4}\,dx = 2\ln|x^2 - 4| + K$
If a quadratic factor with positive discriminant → it factors: one logarithm per linear factor

Example: $x^2 - 2x - 3 = (x - 3)(x + 1)$
If a quadratic factor with zero discriminant → a perfect square, that is a repeated factor: it gives a power

Example: $\int \frac{dx}{x^2 - 2x + 1} = -\frac{1}{x - 1} + K$
If a quadratic factor with negative discriminant → complete the square: logarithm plus arctangent

Example: $\int \frac{dx}{x^2 - 2x + 5} = \frac{1}{2}\arctan\frac{x - 1}{2} + K$
If a cubic or quartic denominator → grouping, the rational root test, a difference of squares, until every factor is linear or irreducible

Example: $x^3 - 3x + 2 = (x - 1)^2(x + 2)$ , a hidden repeated root
If $\sqrt{x}$ or $e^x$ in the integrand → rationalize: $u = \sqrt{x}$ with $dx = 2u\,du$ , or $u = e^x$ with $dx = \frac{du}{u}$ , then start the tree again

Example: $\int \frac{\sqrt{x}}{x - 4}\,dx$ becomes $\int \frac{2u^2}{u^2 - 4}\,du$ , improper again

The three middle branches are one question, the sign of $b^2 - 4ac$ . Answer it in writing on the first line: it decides between logarithms, a power and an arctangent before any constant is computed.

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.

Integrating a rational function

When to use it: Any question of the form evaluate $\int \frac{N(x)}{D(x)}\,dx$ , definite or not

1 State the degrees. If $\deg N \ge \deg D$ , write the long division with its placeholders and the line $\frac{N}{D} = Q + \frac{R}{D}$ .
2 Factor $D$ completely and justify each quadratic that stays with its discriminant.
3 Write the form with letters, then count the unknowns against $\deg D$ .
4 Clear the denominators into a polynomial identity; use the roots (cover-up) first, then coefficients or extra values for what is left. Check at one more value of $x$ .
5 Integrate term by term, naming the rule of each: $\frac{1}{p}\ln|px + q|$ , the power rule, completing the square and the arctangent. For a definite integral, say that no pole lies in the interval before using the Fundamental Theorem.

Concluding sentence

“Since $\deg N < \deg D$ and $x^2 + 9$ has discriminant $-36 < 0$ , the decomposition has the form $\frac{A}{x} + \frac{Bx + C}{x^2 + 9}$ .”

The trap: Writing the constants without the identity they come from. A marker cannot give method marks to $A = 1$ , $B = 2$ , $C = 2$ appearing from nowhere, and a single arithmetic slip then costs everything.

Marking: Typically 1 to 2 marks for the division and the factorization, 1 for the form, 2 for the constants, 2 to 3 for the integration and 1 for the exact final value.

Check before you hand in

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

Count the unknowns

The number of letters in the form must equal the degree of the denominator. Too few means a missing power; too many means an invented term.

$\frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 4}$ : four letters for $x^2(x^2 + 4)$ , degree $4$ .
Test the decomposition at one extra value

Pick a value of $x$ not used to find the constants, away from the poles, and compare both sides.

$\frac{9}{(x - 1)^2(x + 2)} = -\frac{1}{x - 1} + \frac{3}{(x - 1)^2} + \frac{1}{x + 2}$ at $x = 0$ : $\frac{9}{2}$ against $1 + 3 + \frac{1}{2} = \frac{9}{2}$ .
Differentiate the antiderivative

It catches the forgotten $\frac{1}{p}$ , the logarithm of a square and the logarithm over a quadratic in one go.

$\frac{d}{dx}\left[\frac{1}{2}\ln(x^2 + 4) + \arctan\frac{x}{2}\right] = \frac{x}{x^2 + 4} + \frac{2}{x^2 + 4} = \frac{x + 2}{x^2 + 4}$ .
Sign and size of a definite integral

Before handing in, compare the exact value with the integrand's sign on the interval and with (length) times (a typical value).

$\int_{-1/2}^{1/2} \frac{dx}{x^2 - 1} = -\ln 3 \approx -1.10$ : negative, since $x^2 - 1 < 0$ there, and of size about $1$ on an interval of length $1$ .

The typical problem, taken apart

A repeated factor and an irreducible quadratic in the same integral

Find $\int \frac{2x^3 + 3x^2 + 4x + 4}{x^2(x^2 + 4)}\,dx$ .

No calculator. Every step must be justified as on a MATH 141 final.

Step 1

Degrees $3$ and $4$ : proper, no division. $x^2$ is the linear factor $x$ repeated, $x^2 + 4$ has discriminant $-16 < 0$ . Form: $\frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 4}$ , four unknowns for degree $4$ .

Why

The form is the first mark of the question, and the count of unknowns is its free check. Forgetting $\frac{B}{x^2}$ here would make the system inconsistent.

Step 2

Clearing: $2x^3 + 3x^2 + 4x + 4 = Ax(x^2 + 4) + B(x^2 + 4) + (Cx + D)x^2$ . At $x = 0$ : $4 = 4B$ , so $B = 1$ .

Why

The only real root, $x = 0$ , kills every term but one: the cover-up gives the HIGHEST power of the repeated factor, never the lower one.

Step 3

Coefficient of $x$ : $4 = 4A$ , so $A = 1$ . Coefficient of $x^3$ : $2 = A + C$ , so $C = 1$ . Coefficient of $x^2$ : $3 = B + D$ , so $D = 2$ .

Why

With no other real root, matching coefficients finishes the job; choosing the equations that contain one unknown at a time avoids solving a real system.

Step 4

Integrate: $\int \frac{dx}{x} = \ln|x|$ , $\int \frac{dx}{x^2} = -\frac{1}{x}$ , and $\frac{x + 2}{x^2 + 4} = \frac{1}{2} \cdot \frac{2x}{x^2 + 4} + \frac{2}{x^2 + 4}$ gives $\frac{1}{2}\ln(x^2 + 4) + \arctan\frac{x}{2}$ .

Why

Each term by its own rule: a logarithm, a power, then the split of the linear numerator around the derivative $2x$ , with $\frac{1}{a} = \frac{1}{2}$ in front of the arctangent and $2 \cdot \frac{1}{2} = 1$ .

Step 5

Check at $x = 1$ : left $\frac{2 + 3 + 4 + 4}{1 \cdot 5} = \frac{13}{5}$ ; right $1 + 1 + \frac{1 + 2}{5} = \frac{13}{5}$ .

Why

One value not used in the computation confirms all four constants at once, in ten seconds.

The conclusion, written out

“ $\int \frac{2x^3 + 3x^2 + 4x + 4}{x^2(x^2 + 4)}\,dx = \ln|x| - \frac{1}{x} + \frac{1}{2}\ln(x^2 + 4) + \arctan\frac{x}{2} + K$ .”

The classic mistake on this problem: Writing $\frac{A}{x^2} + \frac{Cx + D}{x^2 + 4}$ with no $\frac{A}{x}$ term, or integrating $\frac{2}{x^2 + 4}$ into $2\arctan\frac{x}{2}$ or into $\ln(x^2 + 4)$ .

Learn by heart

• Degree check first: $\deg N \ge \deg D$ means long division, and only the remainder is decomposed.
• Factor completely; the discriminant decides every quadratic: positive, logarithms; zero, a power; negative, an arctangent.
• One term per power of each factor; a linear numerator over each irreducible quadratic; unknowns $= \deg D$ .
• $\int \frac{dx}{px + q} = \frac{1}{p}\ln|px + q|$ . $\int \frac{dx}{(x - a)^2} = -\frac{1}{x - a}$ .
• $\int \frac{du}{u^2 + a^2} = \frac{1}{a}\arctan\frac{u}{a}$ , after completing the square and splitting the numerator.
• Cover-up gives the constant of each simple factor and of the highest power of a repeated one; the rest by coefficients.
• $u = \sqrt{x}$ : $dx = 2u\,du$ . $u = e^x$ : $dx = \frac{du}{u}$ .
• Call the constant of integration $K$ and check the answer by differentiating it.

Frequently asked questions

When do I need long division before partial fractions?

Whenever the degree of the numerator is greater than or equal to the degree of the denominator. Partial fractions only decompose a proper fraction. Divide first, keep the polynomial quotient, which integrates directly, and decompose only the remainder over the denominator. Skipping the division loses the quotient, even when the cover-up seems to give sensible constants.

How do I know if a quadratic factor is irreducible?

Compute its discriminant, b squared minus four a c. If it is negative, the quadratic has no real root and stays whole in the decomposition, with a linear numerator on top, and it will give an arctangent. If it is positive, it factors into two linear factors and gives logarithms. If it is zero, it is a perfect square, a repeated factor that gives a power.

Why does a repeated factor need one term for each power?

Because the number of unknown constants must equal the degree of the denominator for the system to have a solution. A squared factor has degree two, so it needs two terms, one over the factor and one over its square. With only the squared term, the fraction cannot blow up at the right rate and the equations contradict each other.

What is the cover-up method and when does it fail?

Multiply the decomposition by one linear factor and evaluate at its root: every other term vanishes and the constant appears. It works for each simple linear factor and for the highest power of a repeated one. It cannot find the lower powers, the numerators over irreducible quadratics, or the missing quotient of an improper fraction; use coefficients or extra values for those.

Why is there an arctangent in some partial fraction integrals?

An irreducible quadratic becomes, after completing the square, a sum of a square and a positive constant. A constant over that sum is the derivative of an arctangent: the integral of one over u squared plus a squared is one over a times the arctangent of u over a. The part of the numerator proportional to the derivative of the quadratic gives a logarithm instead.

Practise it

Corrected exercises: Partial fractions, MATH 141 at McGill

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.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Trigonometric substitution Next sheet Strategy for integration

Looking for a MATH 141 tutor in Montreal?

Get in touch for a first session. Partial fractions are where algebra from high school decides a calculus mark: division, factoring and a small linear system, used on one integral.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: partial fractions (MATH 141)

The essentials

The order that decides the form

Three kinds of term, three kinds of antiderivative

The rules in table form

Each factor, the terms it opens, and what they integrate to

The mistakes that cost marks

1. Decomposing an improper fraction without dividing first

2. Forgetting the inner coefficient of a linear factor

3. Leaving out the lower power of a repeated factor

4. Integrating a squared factor into a logarithm

5. Taking a quadratic that factors for an irreducible one

6. Writing a logarithm over an irreducible quadratic

7. Stopping the factorization of the denominator too early

8. Losing du over u after the substitution u = e^x

Which method to choose

What to do, by the FORM of the rational function

How the answer is expected to be written

Integrating a rational function

Check before you hand in

The typical problem, taken apart

A repeated factor and an irreducible quadratic in the same integral

Learn by heart

Frequently asked questions

When do I need long division before partial fractions?

How do I know if a quadratic factor is irreducible?

Why does a repeated factor need one term for each power?

What is the cover-up method and when does it fail?

Why is there an arctangent in some partial fraction integrals?

Corrected exercises: Partial fractions, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?