MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: strategy for integration (MATH 141)

Q: How do I know which integration technique to use on a MATH 141 exam?

Read the integrand before choosing. First try to simplify it with algebra or an identity. Then look for a function whose derivative is also a factor, and substitute it. Only then name the technique the new form calls for: parts for a product, partial fractions for a rational function, a trigonometric substitution for the square root of a squared constant minus x squared. Most exam integrals take two of these moves.

Q: Why does my substitution not work even though it looks natural?

A substitution needs the derivative of the new variable to be present in the integrand, up to a constant factor. Setting u equal to the denominator only works if its derivative is the numerator. When the derivative is missing, the substitution is only a change of name and the integral does not get simpler: look for another inner function, or simplify first.

Q: Can every function be integrated?

Every continuous function has a definite integral on a closed interval, but not every one has an antiderivative made of the usual functions. The exponential of x squared, the sine of x squared, e to the x over x, and one over the natural log of x are the classic examples. For them you do not search for a formula: you say it has no elementary antiderivative and, if a definite integral is asked, you bound it by comparison.

Q: Why does my answer look different from the answer in the book?

Two antiderivatives of the same function differ by a constant, and different routes often produce answers that look unrelated. One half of sine squared and minus one quarter of cosine of 2x are both correct antiderivatives of sine times cosine, because they differ by one quarter. Differentiate your answer: if you get the integrand back, it is right.

Q: When should I integrate by parts before substituting?

Almost never when the difficulty is a composition, such as e to the square root of x or the cosine of the natural log of x. Substitute the inner function first and rewrite dx completely: this produces a product, which integration by parts then finishes. Parts goes first only when the integrand is already a product of two simple factors.

This sheet is not a summary of section 7.5 of Stewart: you already know the techniques one by one. It answers one question only, what makes students lose marks when the integral on a MATH 141 exam at McGill University comes with no technique announced, and which precise gesture avoids each loss.

Every value below is exact and computed by hand, as on the midterm and the final, and every antiderivative quoted has been checked the only way that settles it: by differentiating it back.

The thread of the chapter

On the exam the integral comes without its chapter: READ it before choosing. Simplify first, then look for a piece whose derivative is a factor, and only then name the technique that the NEW form calls for. The form after the first move decides, never the first glance.

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

Strategy for integration
Algebraic simplification of an integrand
Substitution rule
Integration by parts
Rationalizing substitution
Partial fraction decomposition
Completing the square
Trigonometric substitution
Non-elementary antiderivative
Comparison properties of the integral
Fundamental Theorem of Calculus

The essentials

The four moves of a naked integral, in this order

• 1. SIMPLIFY: expand a square, split a numerator along the denominator, factor a difference of squares, rewrite in $\sin$ and $\cos$ , use $1 + \cos 2x = 2\cos^2 x$ or a conjugate. Then read the integral again.
• 2. LOOK FOR $g$ AND $g'$ : if a function $g(x)$ and, up to a constant factor, its derivative are both in the integrand, set $u = g(x)$ . The test is that the derivative is PRESENT, not that the substitution looks natural.
• 3. CLASSIFY THE FORM that is left: a product (parts), powers of $\sin$ , $\cos$ , $\tan$ , $\sec$ (identities), $\sqrt{a^2 - x^2}$ and its cousins (trigonometric substitution), a rational function (divide, then partial fractions or complete the square), a root of $x$ or of $e^x$ (rationalize).
• 4. TRY AGAIN: most exam integrals take two moves, substitution then parts ( $\int e^{\sqrt{x}}\,dx$ ), rationalizing then long division and partial fractions ( $\int \frac{x\,dx}{x - \sqrt{x} - 2}$ ), substitution then completing the square.
• Every substitution replaces THREE things: the inner function, $dx$ , and the bounds. Every answer is checked by differentiating it.

The first line of a correct solution names the move, with its $dx$ and its bounds. That line is where the method marks are, and it is the line students skip.

No elementary antiderivative does not mean no integral

• $e^{x^2}$ , $e^{-x^2}$ , $\sin(x^2)$ , $\frac{e^x}{x}$ , $\frac{1}{\ln x}$ have no antiderivative made of the usual functions. No technique will find one: recognise them and stop trying.
• The definite integral still EXISTS on any interval where the integrand is continuous, and $F(x) = \int_a^x f(t)\,dt$ still satisfies $F'(x) = f(x)$ .
• It can be BOUNDED by comparison: $1 - x^2 \le e^{-x^2} \le 1$ on $[0, 1]$ gives $\frac{2}{3} \le \int_0^1 e^{-x^2}\,dx \le 1$ .
• Two non-elementary pieces can CANCEL: $\int (1 + 2x^2)e^{x^2}\,dx = xe^{x^2} + C$ , by parts on $\int x \cdot 2xe^{x^2}\,dx$ .

e^{-x^2}

has no elementary antiderivative, yet the shaded area exists: it lies above

y = 1 - x^2

and below

y = 1

, so between

\frac{2}{3}

and

1

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.

Lookalikes: same shape, different first move

Read each line as: this integrand, this first move, this result. Lines that look almost the same call for different tools; the red lines are the moves students invent.

Integral	First move	Antiderivative
$\int xe^{x^2}\,dx$	$u = x^2$ , the $x$ is there	$\frac{1}{2}e^{x^2} + C$
Example: $\int_0^1 xe^{x^2}\,dx = \frac{e - 1}{2}$
$\int x^3e^{x^2}\,dx$	$u = x^2$ , then parts	$\frac{1}{2}(x^2 - 1)e^{x^2} + C$
Example: $\int_0^1 x^3e^{x^2}\,dx = \frac{1}{2}\left[(u - 1)e^u\right]_0^1 = \frac{1}{2}$
$\int e^{x^2}\,dx$	no move works	none no elementary antiderivative
Example: $\frac{4}{3} \le \int_0^1 e^{x^2}\,dx \le \frac{e + 2}{3}$ What to do: Say so, then bound it by comparison: $1 + x^2 \le e^{x^2} \le 1 + (e - 1)x^2$ on $[0, 1]$ .
$\int \frac{x\,dx}{\sqrt{4 - x^2}}$	$u = 4 - x^2$	$-\sqrt{4 - x^2} + C$
Example: $\int_0^1 \frac{x\,dx}{\sqrt{4-x^2}} = 2 - \sqrt{3}$
$\int \frac{dx}{\sqrt{4 - x^2}}$	the table	$\arcsin\frac{x}{2} + C$
Example: $\int_0^1 \frac{dx}{\sqrt{4-x^2}} = \frac{\pi}{6}$
$\int \sqrt{4 - x^2}\,dx$	$x = 2\sin\theta$ , or geometry	$\frac{x}{2}\sqrt{4 - x^2} + 2\arcsin\frac{x}{2} + C$
Example: $\int_0^1 \sqrt{4 - x^2}\,dx = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$
$\int \frac{x\,dx}{x^4 + 1}$	$u = x^2$ , then the table	$\frac{1}{2}\arctan(x^2) + C$
Example: $\int_0^1 \frac{x\,dx}{x^4 + 1} = \frac{1}{2}\arctan 1 = \frac{\pi}{8}$
$\int \frac{dx}{g(x)}$	copy the log rule	$\ln\|g(x)\|$ no such rule
Example: $\frac{d}{dx}\ln(x^4 + 1) = \frac{4x^3}{x^4 + 1}$ , not $\frac{1}{x^4 + 1}$ What to do: Only $\int \frac{g'(x)}{g(x)}\,dx = \ln\|g(x)\| + C$ exists: check that the numerator IS the derivative of the denominator.
$\int e^{g(x)}\,dx$	divide by $g'(x)$	$\frac{e^{g(x)}}{g'(x)}$ no such rule
Example: $\frac{d}{dx}\frac{e^{x^2}}{2x} = e^{x^2} - \frac{e^{x^2}}{2x^2} \neq e^{x^2}$ What to do: Dividing by the inner derivative works only when it is a constant, as in $\int e^{3x}\,dx = \frac{1}{3}e^{3x} + C$ .

Every blue line is confirmed by differentiating its last cell. Every red line fails that same test in one line, which is the fastest way to catch it on your own paper.

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. Substituting when the derivative is not in the integrand

the whole question: the rest of the work is built on a wrong integrand

What not to write

“ $\int \frac{\sec^2 x}{\tan^2 x + 4\tan x + 5}\,dx = \ln(\tan^2 x + 4\tan x + 5) + C$ .”

What to write

“ $u = \tan x$ , $du = \sec^2 x\,dx$ : $\int \frac{du}{(u + 2)^2 + 1} = \arctan(\tan x + 2) + C$ .”

Why: A logarithm needs the derivative of the WHOLE denominator on top, here $(2\tan x + 4)\sec^2 x$ . What is on top is $\sec^2 x$ , the derivative of $\tan x$ : that is the substitution, and the quadratic left over is completed into a square.

2. Renaming the inner function but not the differential

2 to 3 marks out of 4, and the integration by parts that the question was really about

What not to write

“With $t = \sqrt{x}$ : $\int e^{\sqrt{x}}\,dx = \int e^t\,dt = e^{\sqrt{x}} + C$ .”

What to write

“ $t = \sqrt{x}$ , $x = t^2$ , $dx = 2t\,dt$ : $\int 2te^t\,dt = 2(t - 1)e^t$ , so $\int e^{\sqrt{x}}\,dx = 2(\sqrt{x} - 1)e^{\sqrt{x}} + C$ .”

Why: A substitution is not a change of name. $dx = 2t\,dt$ brings in the factor $2t$ , and that factor is what turns the problem into a product for integration by parts. Differentiating $e^{\sqrt{x}}$ gives $\frac{e^{\sqrt{x}}}{2\sqrt{x}}$ and exposes the error in ten seconds.

3. Keeping the old bounds after a substitution

1 to 2 marks, and a value that fails any size check

What not to write

“ $\int_0^4 e^{\sqrt{x}}\,dx = \int_0^4 2te^t\,dt = 6e^4 + 2$ .”

What to write

“ $t = \sqrt{x}$ runs from $0$ to $2$ : $\int_0^2 2te^t\,dt = 2\left[(t - 1)e^t\right]_0^2 = 2(e^2 + 1)$ .”

Why: New variable, new bounds, on the same line as the substitution. The size check catches it: the integrand is at most $e^2 \approx 7.4$ on an interval of length $4$ , so the value is below $30$ , and $6e^4 + 2$ is above $300$ .

4. Writing the square root of a square without absolute value

2 marks, and a zero that should have rung the alarm

What not to write

“ $\int_0^{2\pi}\sqrt{1 - \cos 2x}\,dx = \sqrt{2}\int_0^{2\pi}\sin x\,dx = 0$ .”

What to write

“ $\sqrt{2\sin^2 x} = \sqrt{2}\,|\sin x|$ , so the value is $\sqrt{2}(2 + 2) = 4\sqrt{2}$ .”

\sqrt{1 - \cos 2x}

is the two humps

\sqrt{2}\,|\sin x|

, always above the axis; writing

\sqrt{2}\sin x

flips the second hump below it, and the two areas cancel to a false

0

Why: $\sqrt{a^2} = |a|$ , always. A square root is never negative, so its integral over an interval where it is not identically zero is POSITIVE: an answer of $0$ is impossible before any computation.

5. Splitting into partial fractions before dividing

2 marks, and ten minutes on a system with no solution

What not to write

“After $t = \sqrt{x}$ in $\int \frac{x\,dx}{x - \sqrt{x} - 2}$ : $\frac{2t^3}{(t - 2)(t + 1)} = \frac{A}{t - 2} + \frac{B}{t + 1}$ .”

What to write

“Degrees $3$ and $2$ , so divide first: $\frac{2t^3}{t^2 - t - 2} = 2t + 2 + \frac{6t + 4}{(t - 2)(t + 1)} = 2t + 2 + \frac{16/3}{t - 2} + \frac{2/3}{t + 1}$ .”

Why: Partial fractions only decompose a PROPER fraction: the right-hand side always has numerators of lower degree, so it can never produce the polynomial part $2t + 2$ . Compare the degrees on the line right after the substitution, every time.

6. Clearing one root out of two

the question: the new integrand is still not rational

What not to write

“For $\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}$ , set $t = \sqrt{x}$ , so $\sqrt[3]{x} = t^{2/3}$ .”

What to write

“The indices are $2$ and $3$ , their LCM is $6$ : $t = x^{1/6}$ , $\sqrt{x} = t^3$ , $\sqrt[3]{x} = t^2$ , $dx = 6t^5\,dt$ , and the integrand becomes $\frac{6t^3}{t + 1}$ .”

Why: A rationalizing substitution has to clear EVERY root at once. $t = x^{1/n}$ with $n$ the least common multiple of the indices does it; any smaller choice leaves a fractional power behind. On $[1, 64]$ the value is $11 + 6\ln\frac{2}{3}$ .

7. Dividing by the derivative of the inner function

all the marks for the question, and the time spent defending the answer

What not to write

“ $\int e^{x^2}\,dx = \frac{e^{x^2}}{2x} + C$ .”

What to write

“ $e^{x^2}$ has no elementary antiderivative. On $[0, 1]$ : $\frac{4}{3} \le \int_0^1 e^{x^2}\,dx \le \frac{e + 2}{3}$ .”

Why: Dividing by $g'(x)$ is legal only when $g'$ is a constant ( $\int e^{3x}\,dx = \frac{1}{3}e^{3x}$ ). The quotient rule shows the failure at once: $\frac{d}{dx}\frac{e^{x^2}}{2x} = e^{x^2} - \frac{e^{x^2}}{2x^2}$ .

8. Integrating by parts before substituting a composition

the time of the whole exam question, since each round of parts makes it worse

What not to write

“ $\int e^{\sqrt{x}}\,dx$ : $u = e^{\sqrt{x}}$ , $dv = dx$ , so $xe^{\sqrt{x}} - \frac{1}{2}\int \sqrt{x}\,e^{\sqrt{x}}\,dx$ , and again...”

What to write

“The difficulty is the composition: substitute $t = \sqrt{x}$ first, THEN integrate the product $2te^t$ by parts.”

Why: Parts is a tool for PRODUCTS. Applied to a composition, it differentiates the composite and drags the inner derivative into the new integral. Substituting the inner function is what turns the composition into a product that parts can finish.

9. Reaching for partial fractions when a substitution suffices

no marks if it is finished, but half an hour of an exam and many chances to slip

What not to write

“ $\int \frac{x^3}{x^4 + 1}\,dx$ : I factor $x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$ and decompose.”

What to write

“The numerator is a quarter of the derivative of the denominator: $u = x^4 + 1$ gives $\frac{1}{4}\ln(x^4 + 1) + C$ .”

Why: The factorization is correct and the decomposition would work, with four constants, two logarithms and two arctangents recombining into one logarithm. The test for $g$ and $g$ prime comes BEFORE any decomposition: it costs one line and often ends the question.

Which method to choose

Which first move, by the FORM of the integrand

Before writing any substitution, look at the shape of the integrand and at what sits next to what

\int_0^{\sqrt{2}}\sqrt{4 - x^2}\,dx

is a sector

S

of radius

2

and angle

\frac{\pi}{4}

, area

\frac{\pi}{2}

, plus a triangle

T

of area

1

: no substitution needed.

If the integrand simplifies: a square to expand, a difference of squares, a numerator that splits along the denominator, a quotient of trigonometric functions → simplify, then read the integral again from the top

Example: $\frac{e^{2x} - 1}{e^x + 1} = e^x - 1$ , and $\frac{(x+1)^2}{x^2+1} = 1 + \frac{2x}{x^2+1}$
If a function $g(x)$ and, up to a constant, $g'(x)$ are both factors → $u = g(x)$ , then read the new form

Example: $\frac{1}{x(1 + (\ln x)^2)}$ : $u = \ln x$ gives $\frac{1}{1 + u^2}$ , an arctangent
If a composition $f(\sqrt{x})$ , $f(\ln x)$ or $f(e^x)$ with no $g'$ in sight → substitute the inner function anyway, rewrite $dx$ completely, then usually integrate by parts

Example: $\int_0^1 e^{\sqrt{x}}\,dx = \int_0^1 2te^t\,dt = 2$
If a product, or a lone $\ln$ or $\arctan$ of something → parts, with $u$ the factor that simplifies when differentiated

Example: $\int_0^1 \ln(1 + x^2)\,dx$ : $u = \ln(1 + x^2)$ , $dv = dx$ , then a division, gives $\ln 2 - 2 + \frac{\pi}{2}$
If a rational function → divide if the degrees require it, factor the denominator, then partial fractions or complete the square

Example: $\frac{t}{(t+1)(t+2)} = -\frac{1}{t+1} + \frac{2}{t+2}$
If a root of $x$ or of $e^x$ that blocks everything → set t equal to the WHOLE root, with the LCM of the indices if there are several

Example: $t = \sqrt{e^x - 1}$ turns $\int_0^{\ln 2}\sqrt{e^x - 1}\,dx$ into $\int_0^1 \frac{2t^2}{t^2 + 1}\,dt = 2 - \frac{\pi}{2}$
If $\sqrt{a^2 - x^2}$ in the numerator → trigonometric substitution, or geometry when the integral is a piece of a disc

Example: $\int_0^{\sqrt{2}}\sqrt{4 - x^2}\,dx = \frac{\pi}{2} + 1$ , a sector plus a triangle
If $e^{x^2}$ , $e^{-x^2}$ , $\sin(x^2)$ , $\frac{e^x}{x}$ or $\frac{1}{\ln x}$ alone → no elementary antiderivative: bound the definite integral, or look for a cancellation

Example: $\frac{2}{3} \le \int_0^1 e^{-x^2}\,dx \le 1$

If a first move leaves something no simpler than the start, undo it and take another branch. Two moves are normal on this chapter; four means the first one was wrong.

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.

Writing up an integral that came with no technique

When to use it: Any question that says Evaluate, or Find, followed by an integral and nothing else

1 First line: name the move. For a substitution, write $t$ , $x$ in terms of $t$ if needed, $dx$ in terms of $dt$ , and the new bounds, all on one line. For parts, write $u$ , $dv$ , $du$ , $v$ .
2 Write the new integral in full, then compare the degrees or read the new form before choosing the second move.
3 Name the second technique the same way: the division, the decomposition with its constants, the completed square.
4 Evaluate with exact values only: $\arctan 1 = \frac{\pi}{4}$ , $\ln(e^e) = e$ ; a value like $\arctan 2$ stays as it is.
5 Check: differentiate the antiderivative, or bracket the definite value between the smallest and largest value of the integrand times the length.

Concluding sentence

“Let $t = \sqrt{x}$ , so $x = t^2$ and $dx = 2t\,dt$ ; when $x = 0$ , $t = 0$ and when $x = 4$ , $t = 2$ .”

The trap: Writing $u = \sqrt{x}$ and then using $u$ again for integration by parts. Keep $t$ for the substitution and $u$ , $dv$ for the parts, or the two steps become unreadable to the marker.

Marking: On a typical four-mark integral: 1 mark for the correct first move with its dx and bounds, 1 for the second technique set up correctly, 1 for the antiderivative, 1 for the exact value. A correct number without the moves earns little.

Check before you hand in

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

Differentiate the antiderivative

Thirty seconds, and it catches a forgotten $dx$ , a wrong sign in partial fractions and an invented rule all at once.

$\frac{d}{dx}\left(e^x - \arctan e^x\right) = e^x - \frac{e^x}{1 + e^{2x}} = \frac{e^{3x}}{e^{2x} + 1}$ : consistent.
Bracket the definite value

If the integrand is monotonic on the interval, the value lies between its two end values times the length. A wrong sign or a missed bound lands outside.

$\int_9^{16}\frac{x\,dx}{x - \sqrt{x} - 2}$ : between $7 \cdot \frac{8}{5}$ and $7 \cdot \frac{9}{4}$ , and $9 + 4\ln 2 + \frac{2}{3}\ln 5 \approx 12.8$ fits.
Recompute the bounds of every substitution

Plug each old bound into the formula for $t$ , one at a time, on paper.

$t = \ln x$ on $[e, e^e]$ : $t$ runs from $1$ to $\ln(e^e) = e$ , not to $e^{e-1}$ .
Check the sign before the number

A square root or a square is never negative, so its integral is positive; an integrand negative on most of the interval gives a negative value.

$\int_0^{2\pi}\sqrt{1 - \cos 2x}\,dx = 4\sqrt{2} > 0$ , and $\int_0^{\sqrt{\pi}} x^3\cos(x^2)\,dx = -1 < 0$ .

The typical problem, taken apart

A substitution, then a rational function read correctly

Evaluate $\displaystyle\int_1^{e} \frac{\ln x}{x(1 + \ln x)^2}\,dx$ exactly.

No calculator, no technique announced: every move must be named, as on a MATH 141 final.

The integrand is positive on

[1, e]

, so the value must be positive; the shaded area is what the exact answer

\ln 2 - \frac{1}{2}

measures.

Step 1

Read the integrand: $\ln x$ appears, and $\frac{1}{x}$ is a factor. Set $t = \ln x$ , $dt = \frac{dx}{x}$ ; bounds $1 \mapsto 0$ and $e \mapsto 1$ . The integral becomes $\int_0^1 \frac{t}{(1 + t)^2}\,dt$ .

Why

The derivative of $\ln x$ is PRESENT, which is the test for a substitution. Changing the bounds now means never going back to $x$ .

Step 2

Read the new form: a proper rational function (degree $1$ over degree $2$ ) with a repeated linear factor. Split the numerator along the denominator: $\frac{t}{(1+t)^2} = \frac{(1 + t) - 1}{(1 + t)^2} = \frac{1}{1 + t} - \frac{1}{(1 + t)^2}$ .

Why

This is the decomposition $\frac{A}{1+t} + \frac{B}{(1+t)^2}$ with $A = 1$ , $B = -1$ , obtained in one line by simplification instead of a system. Either way the method mark is for naming the form.

Step 3

Integrate term by term: $\int \left(\frac{1}{1 + t} - \frac{1}{(1 + t)^2}\right)dt = \ln(1 + t) + \frac{1}{1 + t} + C$ .

Why

The second term is a power, $(1 + t)^{-2}$ , whose antiderivative is $-(1 + t)^{-1}$ : the minus signs cancel. Mixing it up with a logarithm is the usual slip on a repeated factor.

Step 4

Evaluate: $\left[\ln(1 + t) + \frac{1}{1 + t}\right]_0^1 = \left(\ln 2 + \frac{1}{2}\right) - (0 + 1) = \ln 2 - \frac{1}{2}$ .

Why

Exact value, no decimal: on the exam $\ln 2 - \frac{1}{2}$ IS the answer.

Step 5

Check the size: $(1 + t)^2 \ge 4t$ because $(1 - t)^2 \ge 0$ , so $0 \le \frac{t}{(1 + t)^2} \le \frac{1}{4}$ on $[0, 1]$ and the value lies in $[0, \frac{1}{4}]$ . With $\ln 2 \approx 0.69$ : $0.69 - 0.5 = 0.19$ . It fits.

Why

A bracket in one line. A sign slip in step 3, giving $\ln 2 + \frac{1}{2} - 1$ with the wrong sign on the fraction, $\ln 2 - \frac{3}{2} < 0$ , would fail it immediately.

The conclusion, written out

“With $t = \ln x$ , $\int_1^{e}\frac{\ln x}{x(1 + \ln x)^2}\,dx = \int_0^1 \frac{t}{(1 + t)^2}\,dt = \left[\ln(1 + t) + \frac{1}{1 + t}\right]_0^1 = \ln 2 - \frac{1}{2}$ .”

The classic mistake on this problem: Integrating by parts from the start with $u = \ln x$ : the $dv$ left over, $\frac{dx}{x(1 + \ln x)^2}$ , still hides the same substitution, and the computation doubles in length. Or forgetting to change the bounds and evaluating $\ln(1 + t) + \frac{1}{1+t}$ between $1$ and $e$ .

Learn by heart

• Order of moves: simplify, look for $g$ and $g'$ , classify the form, try again. The form after the first move decides.
• Substitution: the derivative must be PRESENT. Replace the inner function, $dx$ and the bounds.
• Composition with $\sqrt{x}$ , $\ln x$ or $e^x$ : substitute first, then parts on the product.
• Rational function: degrees first, divide, then factor and split, or complete the square.
• Several roots: $t = x^{1/n}$ with $n$ the LCM of the indices. A root of $e^x - 1$ : $t$ is the whole root.
• $\sqrt{a^2} = |a|$ . $\int \frac{dx}{g(x)} \neq \ln|g(x)|$ . $\int e^{g(x)}\,dx \neq \frac{e^{g(x)}}{g'(x)}$ .
• No elementary antiderivative: $e^{\pm x^2}$ , $\sin(x^2)$ , $\frac{e^x}{x}$ , $\frac{1}{\ln x}$ . The integral still exists: bound it.
• Check every answer by differentiating it, and every definite value by a bracket.

Frequently asked questions

How do I know which integration technique to use on a MATH 141 exam?

Read the integrand before choosing. First try to simplify it with algebra or an identity. Then look for a function whose derivative is also a factor, and substitute it. Only then name the technique the new form calls for: parts for a product, partial fractions for a rational function, a trigonometric substitution for the square root of a squared constant minus x squared. Most exam integrals take two of these moves.

Why does my substitution not work even though it looks natural?

A substitution needs the derivative of the new variable to be present in the integrand, up to a constant factor. Setting u equal to the denominator only works if its derivative is the numerator. When the derivative is missing, the substitution is only a change of name and the integral does not get simpler: look for another inner function, or simplify first.

Can every function be integrated?

Every continuous function has a definite integral on a closed interval, but not every one has an antiderivative made of the usual functions. The exponential of x squared, the sine of x squared, e to the x over x, and one over the natural log of x are the classic examples. For them you do not search for a formula: you say it has no elementary antiderivative and, if a definite integral is asked, you bound it by comparison.

Why does my answer look different from the answer in the book?

Two antiderivatives of the same function differ by a constant, and different routes often produce answers that look unrelated. One half of sine squared and minus one quarter of cosine of 2x are both correct antiderivatives of sine times cosine, because they differ by one quarter. Differentiate your answer: if you get the integrand back, it is right.

When should I integrate by parts before substituting?

Almost never when the difficulty is a composition, such as e to the square root of x or the cosine of the natural log of x. Substitute the inner function first and rewrite dx completely: this produces a product, which integration by parts then finishes. Parts goes first only when the integrand is already a product of two simple factors.

Practise it

Corrected exercises: Strategy for integration, 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

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Revision sheet: strategy for integration (MATH 141)

The essentials

The four moves of a naked integral, in this order

No elementary antiderivative does not mean no integral

The rules in table form

Lookalikes: same shape, different first move

The mistakes that cost marks

1. Substituting when the derivative is not in the integrand

2. Renaming the inner function but not the differential

3. Keeping the old bounds after a substitution

4. Writing the square root of a square without absolute value

5. Splitting into partial fractions before dividing

6. Clearing one root out of two

7. Dividing by the derivative of the inner function

8. Integrating by parts before substituting a composition

9. Reaching for partial fractions when a substitution suffices

Which method to choose

Which first move, by the FORM of the integrand

How the answer is expected to be written

Writing up an integral that came with no technique

Check before you hand in

The typical problem, taken apart

A substitution, then a rational function read correctly

Learn by heart

Frequently asked questions

How do I know which integration technique to use on a MATH 141 exam?

Why does my substitution not work even though it looks natural?

Can every function be integrated?

Why does my answer look different from the answer in the book?

When should I integrate by parts before substituting?

Corrected exercises: Strategy for integration, MATH 141 at McGill

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Looking for a MATH 141 tutor in Montreal?