Back to the on-screen lesson ·
A decision procedure in order of cost: simplify, look for a spare derivative, then products, trigonometric powers, quadratic radicals and rational functions.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to look at an integrand and name the technique it calls for without evaluating it, ask the questions in order of cost so that an algebraic simplification or a spare derivative is found before anything more elaborate is tried, tell apart integrands that differ by a single factor and want different techniques, notice when a rational function needs no decomposition and when a quadratic radical needs no trigonometry, and check any answer by differentiating it.
You now have substitution, parts, reduction formulas, the parity rules for trigonometric powers, trigonometric substitution, completing the square and partial fractions. Each was taught with integrands chosen to suit it, so the choice never arose: the lesson was the answer.
Outside a lesson nothing announces itself. An integral arrives with no label, and the difference between a two-line solution and a two-page one is almost always which technique was chosen in the first ten seconds. That choice is what this lesson is about, and it is a separate skill from any of the techniques it chooses between.
The shape of an integrand is what remains when you ignore the particular functions: a product of unlike things, a composition with a spare derivative, a quotient of polynomials, a root of a quadratic, a power of a trigonometric function.
A signal is a visible feature that names a technique: a spare factor that is the derivative of something inside, a factorable denominator, a lone radical.
The obstacle is the single thing standing between you and an antiderivative — the root, the product, the denominator. The technique to choose is the one whose first move removes the obstacle rather than rearranging around it.
Ask these questions in this order, and stop at the first yes.
0. Can algebra finish it? Divide an improper rational function. Expand a small power. Split a fraction with several terms on top. This costs one line and sometimes removes the integral entirely.
1. Is the derivative of something inside already present? Then substitute for that something. This is the cheapest technique there is, and it is the one most often missed because the integrand also looks like something else.
2. Is it a product of unlike functions? Then parts, with $u$ the factor that simplifies when differentiated: logarithm, then inverse trigonometric, then polynomial. If the result is the same shape smaller, you have a reduction formula; if it is the same shape again, you have a loop.
3. Is it a power of a trigonometric function? Then the parity rules: peel an odd factor and substitute, or halve the powers with the double-angle identities.
4. Is there a root of a quadratic, alone? Complete the square if necessary, then the trigonometric substitution matching the signs. Alone matters — a spare odd power of $x$ outside the root sends you back to question 1.
5. Is it rational? Then divide if improper, factor the denominator, decompose, integrate the pieces.
6. Is there a root of a linear expression? Substitute for the root.
And finally: differentiate your answer. It is the only check that does not involve redoing the integral, and it takes a few seconds.
Another way: picture
Imagine the six techniques as doors in a corridor, with the cheapest nearest. You walk the corridor in order and take the first door that opens. Almost every wasted hour in this subject comes from starting halfway down the corridor because an integrand looked impressive, and walking past a door that would have opened.
Another way: steps
$\displaystyle\int \frac{x}{\sqrt{9 - x^2}}dx$ and $\displaystyle\int \frac{dx}{\sqrt{9 - x^2}}$. These differ by one factor of $x$ and by a whole technique. The first has $x\,dx$, which is $-\tfrac12 d(9 - x^2)$: substitute, two lines. The second has no spare factor at all: $x = 3\sin\theta$, and the answer is an arcsine. The radical is identical in both; the signal is the factor outside it.
$\displaystyle\int x\ln x\,dx$ and $\displaystyle\int \frac{\ln x}{x}dx$. Again one factor apart. The first is an unlike product: parts, with $u = \ln x$. The second contains $\dfrac{dx}{x}$, which is $d(\ln x)$: substitute $u = \ln x$ and it is $\tfrac12 u^2$. Parts would work on the second too, and takes four times as long.
$\displaystyle\int \frac{x^2}{x^2 + 1}dx$. Rational, so partial fractions — except that it is improper, and question 0 catches it: $\dfrac{x^2}{x^2+1} = 1 - \dfrac{1}{x^2+1}$, giving $x - \arctan x$ immediately. No decomposition was needed, and the denominator does not factor over the reals anyway.
Starting at question 4 because the integrand looks hard. Difficulty of appearance is uncorrelated with which technique applies. Start at question 0 every time.
Missing a spare derivative because the integrand also reads as a product. $2x(x^2+5)^7$ is a product and a composition with its derivative present. The composition wins; it always does.
Choosing parts on an integrand with no product at all. $\int \sec^2 x\,dx$ is a standard form. Parts on it produces a correct and useless identity.
Trigonometric substitution on a radical that has a spare $x$. It works and costs five extra lines plus a triangle.
Decomposing an improper rational function. Question 0 exists partly to catch this.
Never checking. Differentiating the answer catches sign errors, lost constants and dropped factors in seconds, and it is the one step of the whole procedure that nobody skips twice.
Because each technique was taught with its own exercises, the impression left is that integration is a collection of procedures to be executed, and that competence means executing them accurately. Accuracy matters, but it is not where the difficulty lives. A learner who can carry out all six techniques flawlessly and cannot choose between them will spend most of an examination doing correct work on the wrong road.
The evidence is in pairs of integrals that differ by a single factor. $\int \frac{x\,dx}{\sqrt{9-x^2}}$ and $\int \frac{dx}{\sqrt{9-x^2}}$ look almost identical and want different techniques and produce unrelated answers — an algebraic function and an arcsine. Nothing about the arithmetic distinguishes them; only the reading does.
So the practice worth doing is not more integrals. It is looking at integrals and saying what technique they want, without evaluating them. Ten of those in five minutes builds the skill that the hour spent on one integral does not — and it is the reason this lesson asks so many questions it never makes you finish.
$\displaystyle\int x e^{x^2}dx$: is a derivative present? $x\,dx = \tfrac12 d(x^2)$, yes. Substitute $u = x^2$: the answer is $\tfrac12 e^{x^2} + C$.
Question 1 answers it.
$\displaystyle\int x e^{x}dx$: is a derivative present? $x$ is not the derivative of anything inside the exponential. Question 1 fails.
The same-looking integrand, a different answer.
Question 2: an unlike product, so parts with $u = x$, giving $xe^x - e^x + C$. One factor of $x$ inside the exponent changed both the technique and the kind of answer — one is elementary in a line, the other needs a genuine method, and $\int e^{x^2}dx$ has no elementary antiderivative at all.
$\displaystyle\int \frac{2x + 3}{x^2 + 3x + 5}dx$: it is rational, so partial fractions comes to mind — but the denominator is irreducible, so there is nothing to decompose.
Question 5 has nothing to offer.
Question 1: is the derivative of the denominator present? $\dfrac{d}{dx}(x^2 + 3x + 5) = 2x + 3$, which is exactly the numerator.
The signal was there all along.
So the answer is $\ln(x^2 + 3x + 5) + C$, in one line. Asking the questions in order would have caught it before partial fractions was ever considered.
Question 0: nothing to divide or split. Question 1: the numerator is $1$, not the derivative $2x + 4$ of the denominator.
Walk the corridor in order.
Question 5: rational — but the discriminant $16 - 32$ is negative, so the denominator is irreducible and there is nothing to decompose.
A rational function need not need partial fractions.
Complete the square: $(x+2)^2 + 4$, so the integral is $\tfrac12\arctan\dfrac{x+2}{2} + C$. Differentiating it returns the integrand.
Which technique does $\displaystyle\int \sqrt{x^2 + 16}\,dx$ call for?
Match each integral to the technique it calls for.
| An ordinary substitution | Integration by parts | Partial fractions | A trigonometric substitution | |
|---|---|---|---|---|
| $\displaystyle\int 2x(x^{2} + 2)^{4}\,dx$ | ||||
| $\displaystyle\int x^{2}\ln x\,dx$ | ||||
| $\displaystyle\int \dfrac{dx}{(x - 2)(x + 4)}$ | ||||
| $\displaystyle\int \dfrac{dx}{\sqrt{4 - x^{2}}}$ |
Put the steps of deciding how to integrate $\dfrac{x^{2} + 2}{x + 1}$ into the order you do them.
Number the steps in order (write the number in the box):
Say which technique each integrand calls for.
| Technique | |
|---|---|
| $x^{4}e^{x}$ | |
| $\dfrac{4x^{3}}{x^{4} + 2}$ | |
| $\dfrac{x + 2}{(x - 1)(x - 3)}$ |
Evaluate $\displaystyle\int_0^{2} 2x\,(x^{2} + 1)^{2}\,dx$.
Answer:
For $\displaystyle\int \dfrac{x^{3}}{\sqrt{49 - x^{2}}}\,dx$, both a trigonometric substitution and an ordinary one are available. Which is shorter, and why?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Evaluate $\displaystyle\int_0^{2}(5x + 4)^{2}\,dx$.
Answer:
You can read an integrand's shape and name its technique on sight. Say in your own words why $\int x(9-x^2)^{-1/2}dx$ and $\int (9-x^2)^{-1/2}dx$ want different techniques. Next: what integrals are for — areas, volumes, lengths and work.
10. Your turn: $\displaystyle\int \frac{dx}{x^{2} + 4x + 8}$, step 3
And the last step is always the check.