Integral Calculus 201-SN3-RE • CEGEP, Montreal

Revision sheet: substitution, integration by parts and partial fractions (201-SN3-RE)

After the immediate antiderivatives, all of the integral calculus in 201-SN3-RE (formerly 201-NYB) fits in three techniques: substitution, which undoes a chain-rule derivative; integration by parts, which undoes a product; and partial fraction decomposition, which cuts a rational function into integrable pieces.

This sheet does not explain the three methods again: it gives the criterion that decides between them and the list of errors that cost marks even when the chosen technique was the right one, namely limits of integration, the boundary term, dividing first and the form of the numerators. It is written for CEGEP students in the Science program (Sciences de la nature); the corrected set of 20 exercises on the same chapter puts each of these moves to work.

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

An integral is not computed, it is RECOGNIZED. The question is never “how do I integrate”, but “which form does this integrand look like”, and each of the three techniques of the chapter answers one precise form.

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

Integration by substitution
Change of limits of integration
Integration by parts
LIATE rule
Tabular method
Partial fraction decomposition
Repeated linear factor
Irreducible quadratic factor
Cover-up method
Arctangent function

The essentials

Substitution: five forms to recognize at a glance

• $\displaystyle\int\frac{f'}{f}\,dx=\ln\left|f\right|$ ; $\displaystyle\int f'f^{n}\,dx=\frac{f^{n+1}}{n+1}$ ; $\displaystyle\int f'e^{f}\,dx=e^{f}$ ; $\displaystyle\int\frac{f'}{\sqrt{f}}\,dx=2\sqrt{f}$ ; $\displaystyle\int f'\cos f\,dx=\sin f$ .
• The test: differentiate the most complicated expression and look for the result ELSEWHERE in the integrand, up to a constant.
• Set $u$ , compute $du=u'\,dx$ , and replace EVERYTHING, including the $dx$ . A $dx$ that survives signals an incomplete substitution.
• DEFINITE integral: either change the limits along with the variable, or go back to $x$ before evaluating. Never a mix of the two.
• The substitution must be MONOTONIC on the interval: $u=x^{2}$ on $[-1, 2]$ is forbidden without splitting into $[-1, 0]$ and $[0, 2]$ .

A substitution that leaves an $x$ standing is not lost: just express that $x$ in terms of $u$ . That is the case of $\int x\sqrt{x+1}\,dx$ , where $x=u-1$ .

Integration by parts: choose $u$ , then stick with it

• $\displaystyle\int u\,dv=uv-\int v\,du$ . On a definite integral, the boundary term $\left[uv\right]_{a}^{b}$ is evaluated at BOTH limits.
• Priority order for $u$ : LIATE, that is Logarithmic, Inverse (trigonometric), Algebraic, Trigonometric, Exponential.
• The logic of LIATE: choose as $u$ what SIMPLIFIES when differentiated, and as $dv$ what integrates without difficulty.
• Tabular method as soon as a polynomial factor eventually differentiates to zero: two columns, derivatives on the left, antiderivatives on the right, alternating signs.
• Cyclic case: $\displaystyle\int e^{ax}\sin(bx)\,dx=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^{2}+b^{2}}$ . Keep the SAME choice of $u$ in both rounds.

The further LEFT, the more the function simplifies when differentiated: that is the one to take as

u

. The further right, the more easily it integrates: that is the one to keep for

dv

The invisible factor $1$ is useful more often than you think: $\int\ln x\,dx$ , $\int\arctan x\,dx$ and $\int\arcsin x\,dx$ are all done by parts with $dv=1\,dx$ .

Partial fractions: four cases, one count

• DIVIDE first if the degree of the numerator is greater than or equal to that of the denominator. Without this division, the decomposition is impossible.
• Distinct linear factor $(x-a)$ : one term $\dfrac{A}{x-a}$ .
• REPEATED linear factor $(x-a)^{n}$ : $n$ terms, one per power, from $\dfrac{A_{1}}{x-a}$ up to $\dfrac{A_{n}}{(x-a)^{n}}$ .
• IRREDUCIBLE quadratic factor $(x^{2}+bx+c)$ : one term $\dfrac{Bx+C}{x^{2}+bx+c}$ , with a numerator of degree $1$ , never a constant.
• The total number of constants ALWAYS equals the degree of the denominator: this is the check that validates the setup before any calculation.

Two antiderivatives never to be confused: $\displaystyle\int\frac{dx}{x^{2}-a^{2}}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|$ whereas $\displaystyle\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\arctan\frac{x}{a}$ . One sign, two different functions.

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. Keeping the old limits after a substitution

the whole question, 4 marks

What not to write

“ $\displaystyle\int_{0}^{2}2x\,e^{x^{2}}dx$ with $u=x^{2}$ : $\displaystyle\int_{0}^{2}e^{u}du=e^{2}-1$ .”

What to write

“The limits change with the variable: $u$ goes from $0$ to $4$ , so $\displaystyle\int_{0}^{4}e^{u}du=e^{4}-1$ .”

Both areas equal

e^{4}-1

: when

x

goes from

0

2

u=x^{2}

goes from

0

4

. The limits change with the variable.

Why: A limit of integration is a value of the VARIABLE, not a number attached to the integral. The perfectly correct alternative is to go back to $x$ before evaluating, but you must choose one of the two routes.

2. Forgetting the $dx$ in the change of variable

2 marks, and an antiderivative off by a constant factor

What not to write

“ $u=3x+1$ , so $\displaystyle\int\cos(3x+1)\,dx=\int\cos u\,du=\sin(3x+1)$ .”

What to write

“ $du=3\,dx$ , so $dx=\dfrac{du}{3}$ and $\displaystyle\int\cos u\,\frac{du}{3}=\frac{\sin(3x+1)}{3}$ .”

Why: The substitution replaces EVERYTHING that contains $x$ , including the differential. The check is free: differentiating the result must give back the integrand.

3. Substituting on an interval where the change of variable is not monotonic

the whole question, 5 marks

What not to write

“ $\displaystyle\int_{-1}^{2}x^{2}\,dx$ with $u=x^{2}$ : $dx=\dfrac{du}{2\sqrt{u}}$ , so the integral equals $\displaystyle\int_{1}^{4}\frac{\sqrt{u}}{2}\,du=\frac{7}{3}$ .”

What to write

“ $u=x^{2}$ is not monotonic on $[-1, 2]$ : it decreases to $0$ then increases again. The direct calculation gives $\left[\dfrac{x^{3}}{3}\right]_{-1}^{2}=3$ , not $\dfrac{7}{3}$ .”

[-1, 2]

, the substitution goes through

0

before rising again: each value of

u

between

0

and

1

corresponds to TWO values of

x

, which rules out the change of variable without splitting.

Why: Writing $dx=\dfrac{du}{2\sqrt{u}}$ assumes $x=+\sqrt{u}$ , which is false on $[-1, 0]$ where $x$ is negative. The substitution then wipes out part of the interval, and the error here reaches $22\%$ .

4. Choosing $u$ against LIATE

no marks lost directly, but an integral harder than the first one

What not to write

“ $\displaystyle\int x\ln x\,dx$ : I set $u=x$ and $dv=\ln x\,dx$ .”

What to write

“Logarithmic comes before Algebraic: $u=\ln x$ and $dv=x\,dx$ , which gives $\dfrac{x^{2}\ln x}{2}-\dfrac{x^{2}}{4}$ .”

Why: Taking $\ln x$ as $dv$ requires already knowing how to integrate it, which is precisely the problem you were trying to solve. LIATE ranks the functions by how easily they differentiate.

5. Evaluating the boundary term at a single limit

2 marks for rigour, and a real error as soon as the lower limit is not zero

What not to write

“ $\displaystyle\int_{0}^{1}xe^{x}dx=xe^{x}-\int_{0}^{1}e^{x}dx=e-\left[e^{x}\right]_{0}^{1}=e-(e-1)=1$ .”

What to write

“The boundary term is evaluated at BOTH limits: $\left[xe^{x}\right]_{0}^{1}=e-0=e$ , then subtract $e-1$ , which gives $1$ . The result is the same here, but the written work is wrong as soon as the lower limit does not vanish.”

Why: The omission goes unnoticed when the lower limit is zero, which is the case in the most commonly taught example. It shows up on $\displaystyle\int_{1}^{2}xe^{x}dx$ , where the boundary term equals $2e^{2}-e$ .

6. Changing the choice of $u$ in the second round of a cyclic case

the whole question, 5 marks, and ten minutes lost

What not to write

“First round with $u=e^{x}$ , second round with $u=\sin x$ : I land back on $I=I$ .”

What to write

“KEEP the same kind of $u$ in both rounds. You then get $I=\ldots-I$ , so $2I=\ldots$ , and the integral follows.”

Why: Swapping roles in the second round undoes exactly what the first one did: the equality you get is true but empty. Keeping the same choice produces, on the contrary, an equation in $I$ .

7. Decomposing an improper fraction without dividing first

the whole question, 5 marks

What not to write

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

What to write

“The degree of the numerator equals that of the denominator: DIVIDE first, which gives $1+\dfrac{1}{x^{2}-1}$ , then decompose only the remainder.”

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

tends to

1

at infinity: this

1

is the quotient of the division, which no term

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

can produce.

Why: Partial fraction decomposition exists only for a PROPER fraction. The system obtained without dividing simply has no solution, which shows in the impossibility of solving it.

8. Writing only one term for a repeated factor

the whole question, 5 marks

What not to write

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

What to write

“A factor of multiplicity $2$ calls for TWO terms: $\dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{(x-2)^{2}}$ . Three constants for a denominator of degree $3$ .”

Why: Counting the constants is the check: the count must equal the degree of the denominator. With two constants for degree $3$ , the system is underdetermined and has no solution.

9. Putting a constant numerator over an irreducible quadratic factor

the whole question, 5 marks

What not to write

“ $\dfrac{3x^{2}+2x+5}{(x-1)(x^{2}+4)}=\dfrac{A}{x-1}+\dfrac{B}{x^{2}+4}$ .”

What to write

“An irreducible quadratic carries a numerator of degree $1$ : $\dfrac{A}{x-1}+\dfrac{Bx+C}{x^{2}+4}$ . Three constants for a denominator of degree $3$ .”

Why: Without the $Bx$ term, you cannot produce the logarithm $\ln(x^{2}+4)$ that appears in the antiderivative. Counting the constants flags the error before you even solve.

10. Applying the cover-up method where it does not work

3 marks, and a system solved the wrong way

What not to write

“I cover $(x-2)$ in $\dfrac{1}{x(x-2)^{2}}$ and evaluate at $2$ to get $B$ .”

What to write

“Cover-up only gives the coefficient of the HIGHEST power, so $C$ , and nothing on an irreducible quadratic. The other constants are found with a system or with particular values.”

Why: Cover-up relies on cancelling a simple factor. On a repeated factor, covering the whole power still leaves a singularity, and on an irreducible quadratic there is no real root at which to evaluate.

11. Confusing the antiderivatives of $\dfrac{1}{x^{2}-a^{2}}$ and $\dfrac{1}{x^{2}+a^{2}}$

the whole question, 4 marks

What not to write

“ $\displaystyle\int\frac{dx}{x^{2}+4}=\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|$ .”

What to write

“With a PLUS, the denominator does not factor: $\displaystyle\int\frac{dx}{x^{2}+4}=\frac{1}{2}\arctan\frac{x}{2}$ . The logarithm is reserved for the MINUS sign.”

With a MINUS (red), two asymptotes: the denominator factors, hence the logarithm. With a PLUS (green), none: the curve stays bounded, hence the arctangent.

Why: A minus sign makes the denominator factorable, so it decomposes into two fractions with logarithms. A plus sign makes it irreducible, and the antiderivative is an arctangent.

Which method to choose

Which technique according to the FORM of the integrand

What you see in the integrand, in this order

If the derivative of one piece appears elsewhere, up to a constant → SUBSTITUTION, setting that piece as $u$

Example: $\displaystyle\int\frac{2x}{x^{2}+1}dx=\ln(x^{2}+1)$
If a PRODUCT of two functions of different kinds → PARTS, with $u$ chosen by LIATE

Example: $\displaystyle\int x\ln x\,dx$ , $\displaystyle\int xe^{x}dx$
If a RATIONAL FUNCTION whose denominator factors → PARTIAL FRACTIONS, after division if the fraction is improper

Example: $\dfrac{3x^{2}+2x+5}{(x-1)(x^{2}+4)}$
If an $n$ -th root of a simple expression → RATIONALIZING substitution: $x=u^{n}$ , or $u^{m}$ with $m$ the LCM if there are several roots

Example: $\displaystyle\int\frac{dx}{\sqrt{x}+\sqrt[3]{x}}$ with $x=u^{6}$
If a single function with no obvious antiderivative, such as $\ln x$ or $\arctan x$ → PARTS with the invisible factor $1$ , so $dv=1\,dx$

Example: $\displaystyle\int\arctan x\,dx=x\arctan x-\tfrac{1}{2}\ln(1+x^{2})$
If no form stands out → try a simple substitution to SIMPLIFY, then go through the tree again from the top

many exam integrals call for a substitution FOLLOWED BY an integration by parts

The reading order matters: look for a substitution first, because it is the fastest, and move on to parts only if no derivative appears. Partial fractions come last, because they are the longest.

Which form of term for each factor of the denominator

The complete factorization of the denominator, after any division

If distinct linear factor $(x-a)$ → $\dfrac{A}{x-a}$ , one constant, and $A$ is found by cover-up

Example: $\dfrac{2}{x-1}$
If repeated linear factor $(x-a)^{n}$ → $n$ terms, from $\dfrac{A_{1}}{x-a}$ to $\dfrac{A_{n}}{(x-a)^{n}}$

Example: $\dfrac{A}{x-2}+\dfrac{B}{(x-2)^{2}}$
If irreducible quadratic factor $(x^{2}+bx+c)$ → $\dfrac{Bx+C}{x^{2}+bx+c}$ , numerator of degree $1$

Example: $\dfrac{x+3}{x^{2}+4}$
If REPEATED irreducible quadratic factor → one term $\dfrac{B_{k}x+C_{k}}{(x^{2}+bx+c)^{k}}$ for each power

rare at CEGEP, but counting the constants remains the same check
If the degree of the numerator reaches that of the denominator → DIVIDE before anything else, and decompose only the remainder

the quotient is integrated separately, usually as a polynomial

Count the constants BEFORE solving: their number must equal the degree of the denominator. This check alone catches the two most frequent setup errors.

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: The integrand is a quotient of two polynomials.

1 Compare the degrees. If the numerator reaches the degree of the denominator, carry out polynomial long DIVISION and write the integrand as quotient plus remainder over divisor.
2 Factor the denominator completely into linear and irreducible quadratic factors.
3 Set up the decomposition with the right form for each factor, and CHECK that the number of constants equals the degree of the denominator.
4 Bring everything to a common denominator, match the numerators, and solve with particular values (cover-up method) then by equating the remaining coefficients.
5 Integrate term by term: $\dfrac{A}{x-a}$ gives a logarithm, $\dfrac{Bx}{x^{2}+c}$ a logarithm after substitution, $\dfrac{C}{x^{2}+c}$ an arctangent.
6 Check by differentiating, or by evaluating the decomposition identity at a simple value.

Concluding sentence

“The fraction is proper and the denominator factors as $(x-1)(x^{2}+4)$ : we write $\dfrac{A}{x-1}+\dfrac{Bx+C}{x^{2}+4}$ , that is three constants for a denominator of degree three; equating coefficients gives $A=2$ , $B=1$ and $C=3$ .”

The trap: Split the quadratic term into two pieces before integrating. $\dfrac{x+3}{x^{2}+4}$ does not have a single antiderivative: it is $\dfrac{x}{x^{2}+4}$ , which gives a logarithm, PLUS $\dfrac{3}{x^{2}+4}$ , which gives an arctangent.

Marking: Typically 1 mark for checking that the fraction is proper, 1 for the factorization, 2 for the form of the decomposition, 2 for the constants, 3 for the antiderivatives.

Writing up a substitution on a definite integral

When to use it: The integral has limits and the integrand contains a composite function whose derivative appears.

1 Set $u$ and write $du$ in terms of $dx$ , then isolate $dx$ if needed.
2 CHANGE THE LIMITS: compute $u(a)$ and $u(b)$ , and write them explicitly.
3 Check that the substitution is MONOTONIC on the interval, and say so; otherwise, split the interval.
4 Rewrite the integral entirely in $u$ , with no $x$ left, and evaluate it between the new limits.
5 Conclude without going back to $x$ , since the limits have been changed.

Concluding sentence

“Let $u=x^{2}$ , so $du=2x\,dx$ ; the limits become $u(0)=0$ and $u(2)=4$ , and the substitution is increasing on $[0, 2]$ ; the integral then equals $\displaystyle\int_{0}^{4}e^{u}du=e^{4}-1$ .”

The trap: Changing the limits AND going back to $x$ at the end. The two methods are each correct on their own; mixing them evaluates the antiderivative in $x$ at limits expressed in $u$ .

Check before you hand in

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

Differentiate the result

Any antiderivative is checked by differentiating it: you must get back exactly the integrand. It is the universal check of the chapter and it only takes thirty seconds.

An antiderivative $\sin(3x+1)$ , once differentiated, gives $3\cos(3x+1)$ instead of $\cos(3x+1)$ : the factor $\tfrac{1}{3}$ is missing.
Counting the constants against the degree

Before solving a decomposition, count the constants you set up and compare with the degree of the denominator. The two numbers must match.

Two constants for a denominator of degree $3$ : a repeated factor was written only once.
The decomposition identity tested at one point

Substitute a simple value, such as $0$ or $2$ , into the original fraction and into the decomposition you found: both must give the same number.

$-1.25$ on both sides at $x=0$ : the decomposition is right.
Consistent limits after a substitution

Reread the limits of the final integral and check that they are expressed in the SAME variable as the integrand.

An integral in $u$ evaluated between $0$ and $2$ while $u(2)=4$ : the limits did not follow.
Sign and order of magnitude of a definite integral

An integrand that is positive on the interval must give a positive integral. Comparing the result with the area of a rough rectangle gives an order of magnitude.

A negative result for $\displaystyle\int_{0}^{2}2xe^{x^{2}}dx$ , whose integrand is positive: a sign was lost.

The typical problem, taken apart

A rational function with an irreducible quadratic

Compute $\displaystyle\int\frac{3x^{2}+2x+5}{(x-1)(x^{2}+4)}\,dx$ .

Show in detail the setup of the decomposition, how the constants are found and the integration of each term.

The vertical asymptote at

x=1

announces the term

\dfrac{2}{x-1}

, hence a logarithm. The denominator

x^{2}+4

, on the other hand, never vanishes: it adds no asymptote and will give an arctangent.

Step 1

Degree of the numerator: $2$ . Degree of the denominator: $3$ . The fraction is PROPER, no division is needed.

Why

This is the first check, and it takes two seconds. Skipping it on an improper fraction makes the next system unsolvable, with nothing to warn you before the end.

Step 2

The denominator is already factored: a linear factor $(x-1)$ and a quadratic $x^{2}+4$ , irreducible since its discriminant is $-16$ .

Why

Justifying irreducibility in one line saves you from looking for a factorization that does not exist, and above all it dictates the form of the numerator at the next step.

Step 3

Decomposition: $\dfrac{3x^{2}+2x+5}{(x-1)(x^{2}+4)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^{2}+4}$ . Three constants for a denominator of degree $3$ : the count is consistent.

Why

The numerator of the quadratic term has degree $1$ , not a constant. It is this $Bx$ term that will produce the logarithm $\ln(x^{2}+4)$ , impossible to get otherwise.

Step 4

Over a common denominator: $3x^{2}+2x+5=A(x^{2}+4)+(Bx+C)(x-1)$ . At $x=1$ : $10=5A$ , so $A=2$ .

Why

The cover-up method gives $A$ immediately, because $x=1$ cancels the second term. It is the only constant it can supply here, the quadratic having no real root.

Step 5

Equating coefficients: for $x^{2}$ , $3=A+B$ , so $B=1$ ; for $x^{0}$ , $5=4A-C$ , so $C=8-5=3$ . Check with $x^{1}$ : $2=C-B=3-1=2$ , consistent.

Why

Two coefficients are used to find the unknowns and the third to CHECK. It is free, and it replaces a check you would have to do anyway.

Step 6

So $\dfrac{3x^{2}+2x+5}{(x-1)(x^{2}+4)}=\dfrac{2}{x-1}+\dfrac{x+3}{x^{2}+4}$ . Check at $x=0$ : the fraction equals $\dfrac{5}{-4}=-1.25$ , and the decomposition $-2+\dfrac{3}{4}=-1.25$ .

Why

One check point is enough to validate the whole decomposition, and $x=0$ is almost always the fastest to evaluate on both sides.

Step 7

SPLIT the quadratic term: $\dfrac{x+3}{x^{2}+4}=\dfrac{x}{x^{2}+4}+\dfrac{3}{x^{2}+4}$ , because these two pieces do not have the same antiderivative.

Why

This is the step that exam papers skip. The first piece calls for a substitution $u=x^{2}+4$ , the second for an arctangent: handling them together is impossible.

Step 8

Integration: $\displaystyle\int\frac{2\,dx}{x-1}=2\ln\left|x-1\right|$ ; $\displaystyle\int\frac{x\,dx}{x^{2}+4}=\tfrac{1}{2}\ln(x^{2}+4)$ ; $\displaystyle\int\frac{3\,dx}{x^{2}+4}=\tfrac{3}{2}\arctan\dfrac{x}{2}$ .

Why

The second piece has no absolute value, $x^{2}+4$ being always positive. The third uses $\int\dfrac{dx}{x^{2}+a^{2}}=\dfrac{1}{a}\arctan\dfrac{x}{a}$ with $a=2$ , hence the $\tfrac{1}{2}$ multiplied by $3$ .

Step 9

Check by differentiating at $x=0$ : the derivative of $2\ln\left|x-1\right|+\tfrac{1}{2}\ln(x^{2}+4)+\tfrac{3}{2}\arctan\dfrac{x}{2}$ equals $\dfrac{2}{-1}+0+\dfrac{3}{4}=-1.25$ , the same as the integrand.

Why

A single differentiation point validates all three antiderivatives at once. It is the most profitable check in all of integral calculus, and it requires no tools.

The conclusion, written out

“ $\displaystyle\int\frac{3x^{2}+2x+5}{(x-1)(x^{2}+4)}\,dx=2\ln\left|x-1\right|+\frac{1}{2}\ln(x^{2}+4)+\frac{3}{2}\arctan\frac{x}{2}+C$ .”

The classic mistake on this problem: Writing $\dfrac{B}{x^{2}+4}$ with a constant numerator. The system then becomes inconsistent, and if you force it, the antiderivative you get contains no logarithm of the quadratic: its derivative does not give back the integrand.

Learn by heart

• Substitution: replace EVERYTHING, $dx$ included, and change the limits OR go back to $x$ , never both.
• A substitution must be MONOTONIC on the interval, otherwise split it.
• Parts: $\displaystyle\int u\,dv=uv-\int v\,du$ , with $u$ chosen by LIATE and the boundary term evaluated at BOTH limits.
• Cyclic case: keep the same choice of $u$ in both rounds to get an equation in $I$ .
• Partial fractions: DIVIDE if the fraction is improper, then one term per factor.
• Repeated factor of multiplicity $n$ : $n$ terms. Irreducible quadratic: numerator $Bx+C$ .
• The number of constants equals the DEGREE of the denominator: check it before any calculation.
• $\displaystyle\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\arctan\frac{x}{a}$ , whereas $\displaystyle\int\frac{dx}{x^{2}-a^{2}}$ gives a logarithm.

Frequently asked questions

Do you have to change the limits of integration after a substitution?

Yes, if you want to evaluate directly in the new variable. A limit of integration is a value of the variable of integration: when the variable changes, the limits must follow. The correct alternative is to find the antiderivative, go back to the original variable, then evaluate at the original limits. What you must never do is mix the two.

How do you choose u in integration by parts?

By following the LIATE rule: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. Choose as u the function that appears furthest to the left in this list, because it is the one that simplifies the most when differentiated. The rest becomes dv, and it must be easy to integrate.

Why do you have to divide before decomposing a rational function?

Because partial fraction decomposition exists only for a proper fraction, that is one whose numerator has a degree strictly less than that of the denominator. If that is not the case, the system of equations you get simply has no solution. Long division separates a polynomial, easy to integrate, from a proper remainder.

What numerator goes over an irreducible quadratic factor?

A numerator of degree one, of the form B times x plus C, never a mere constant. The x term is essential: it is what produces the logarithm of the quadratic in the antiderivative. Without it, the decomposition cannot reproduce the original fraction and the system becomes inconsistent.

How do you check an antiderivative quickly?

By differentiating it. You must get back exactly the integrand. It is the only universal check in integral calculus, it requires no tools and it catches a forgotten constant factor just as well as a reversed sign. On a decomposition, you can also evaluate both sides at a simple point such as zero.

When is a substitution not allowed?

When it is not monotonic on the interval of integration. Setting u equal to x squared on an interval that contains zero makes two values of x correspond to each value of u, which distorts the change of variable. You must then split the interval at the points where the substitution changes direction.

Practise it

Corrected exercises: Substitution, parts and partial fractions: 20 exercises, 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.

20 corrected exercises
200 points
300 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-substitution-parts-partial-fractions. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: substitution, integration by parts and partial fractions (201-SN3-RE)

The essentials

Substitution: five forms to recognize at a glance

Integration by parts: choose uuu, then stick with it

Partial fractions: four cases, one count

The mistakes that cost marks

1. Keeping the old limits after a substitution

2. Forgetting the dxdxdx in the change of variable

3. Substituting on an interval where the change of variable is not monotonic

4. Choosing uuu against LIATE

5. Evaluating the boundary term at a single limit

6. Changing the choice of uuu in the second round of a cyclic case

7. Decomposing an improper fraction without dividing first

8. Writing only one term for a repeated factor

9. Putting a constant numerator over an irreducible quadratic factor

10. Applying the cover-up method where it does not work

11. Confusing the antiderivatives of 1x2−a2\dfrac{1}{x^{2}-a^{2}}x2−a21​ and 1x2+a2\dfrac{1}{x^{2}+a^{2}}x2+a21​

Which method to choose

Which technique according to the FORM of the integrand

Which form of term for each factor of the denominator

How the answer is expected to be written

Integrating a rational function

Writing up a substitution on a definite integral

Check before you hand in

The typical problem, taken apart

A rational function with an irreducible quadratic

Learn by heart

Frequently asked questions

Do you have to change the limits of integration after a substitution?

How do you choose u in integration by parts?

Why do you have to divide before decomposing a rational function?

What numerator goes over an irreducible quadratic factor?

How do you check an antiderivative quickly?

When is a substitution not allowed?

Corrected exercises: Substitution, parts and partial fractions: 20 exercises, 201-SN3-RE

See also

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Integration by parts: choose $u$ , then stick with it

2. Forgetting the $dx$ in the change of variable

4. Choosing $u$ against LIATE

6. Changing the choice of $u$ in the second round of a cyclic case

11. Confusing the antiderivatives of $\dfrac{1}{x^{2}-a^{2}}$ and $\dfrac{1}{x^{2}+a^{2}}$