Choosing a Technique

Every earlier page in this chapter named its technique before it showed you an integral. Here the problems come mixed, and before the answer box opens you say what you would do first. The page has already tried all five first moves on the integral in front of you: it tells you whether yours reaches the table within three moves, and whether anything gets there sooner.

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Lesson

The theory — Choosing a Technique

Choosing a technique is deciding the first move on an integral that arrives without a label: the table as it stands, a substitution, integration by parts, partial fractions, or a trigonometric rewrite. On this page a move works when, followed by at most two more, it leaves only integrals in the table. Moves are counted as techniques applied: integrating dv by a substitution inside parts is a move of its own, and splitting a sum into its terms is free.

Assumes: The search knows those five moves and nothing else. It cannot complete a square, so ∫dx/(x² + 2x + 5) would be reported as reaching nothing, and none like it is set here. Its trigonometric move is what the previous page teaches: an odd power gives up one factor for du, even powers take the half-angle identity, tan² becomes sec² − 1, secⁿ is reduced, and √(a² − x²), √(x² + a²), √(x² − a²) take x = a sin u, a tan u, a sec u.

Problems solved in full

The top is the derivative of the bottom 2 steps
Find ∫x/(x² + 9) dx.
1. Cue: the numerator is half the derivative of the denominator. Move: substitute the denominator.
  
  $\frac{d}{d x} (x^{2} + 9) = 2 x \Rightarrow u = x^{2} + 9, x d x = \frac{d u}{2}$
2. A logarithm, with no bars because x² + 9 is never negative:
  
  $\int \frac{x}{x ^{2} + 9} d x = \frac{1}{2} \int \frac{d u}{u} = \frac{1}{2} ln (x^{2} + 9) + C$
Answer

½ ln(x² + 9) + C. Partial fractions are no use here: x² + 9 has no real root to split at.

$\int \frac{x}{x ^{2} + 9} d x = \frac{1}{2} ln (x^{2} + 9) + C$
A power of x times an exponential 2 steps
Find ∫x e^3x dx.
Show solution
1. Cue: one factor gets simpler when differentiated and the other is easy to integrate. Move: parts, with u = x.
  
  $u = x, d v = e^{3 x} d x, v = \frac{1}{3} e^{3 x}$
2. One round, and the integral left is in the table:
  
  $\int x e^{3 x} d x = \frac{1}{3} x e^{3 x} - \frac{1}{3} \int e^{3 x} d x = \frac{1}{3} x e^{3 x} - \frac{1}{9} e^{3 x} + C$
Answer

e^3x(x/3 − 1/9) + C. No substitution helps: the x beside e^3x is not the derivative of 3x.

$\int x e^{3 x} d x = e^{3 x} (\frac{x}{3} - \frac{1}{9}) + C$
A denominator that factors 2 steps
Find ∫5/((x + 1)(x − 4)) dx.
Show solution
1. Cue: a rational function whose denominator is already in factors. Move: partial fractions, by cover-up.
  
  $\frac{5}{( x + 1 ) ( x - 4 )} = \frac{A}{x - 4} + \frac{B}{x + 1}, A = \frac{5}{5} = 1, B = \frac{5}{- 5} = - 1$
2. Two logarithms:
  
  $\int (\frac{1}{x - 4} - \frac{1}{x + 1}) d x = ln ∣ x - 4∣ - ln ∣ x + 1∣ + C$
Answer

ln|(x − 4)/(x + 1)| + C. The 5 in the numerator is what the cover-up divides: 5/(4 + 1) and 5/(−1 − 4).

$\int \frac{5 d x}{( x + 1 ) ( x - 4 )} = ln \frac{x - 4}{x + 1} + C$
An even power of cosine 2 steps
Find ∫cos²x dx.
Show solution
1. Cue: an even power, and no odd factor to spend on du. Move: halve the angle.
  
  $cos^{2} x = \frac{1}{2} (1 + cos 2 x)$
2. Integrate term by term:
  
  $\int \frac{1}{2} (1 + cos 2 x) d x = \frac{x}{2} + \frac{sin 2 x}{4} + C$
Answer

x/2 + sin 2x/4 + C. u = sin x does not work: one cos x becomes du, and the other is left as √(1 − u²), because the power is even.

$\int cos^{2} x d x = \frac{x}{2} + \frac{sin 2 x}{4} + C$
It looks like parts and is a substitution 2 steps
Find ∫(ln x)/x dx for x > 0.
Show solution
1. Cue: 1/x is the derivative of ln x, which stands beside it. Move: substitute u = ln x.
  
  $u = ln x, d u = \frac{d x}{x}$
2. A power of u:
  
  $\int \frac{ln x}{x} d x = \int u d u = \frac{u ^{2}}{2} = \frac{( ln x ) ^{2}}{2} + C$
Answer

(ln x)²/2 + C. Parts would work too, the integral coming back as itself, I = (ln x)² − I; the substitution gets there in one line.

$\int \frac{ln x}{x} d x = \frac{( ln x ) ^{2}}{2} + C$
No real roots: complete the square 2 steps
Find ∫dx/(x² − 6x + 10).
Show solution
1. Cue: the discriminant, 36 − 40, is negative, so nothing factors. Move: complete the square.
  
  $36 - 40 < 0; x^{2} - 6 x + 10 = (x - 3)^{2} + 1$
2. The arctangent form, shifted by 3:
  
  $\int \frac{d x}{( x - 3 ) ^{2} + 1} = arctan (x - 3) + C$
Answer

arctan(x − 3) + C. This is the one move the page’s search cannot make, which is why it is worth doing by hand once.

$\int \frac{d x}{x ^{2} - 6 x + 10} = arctan (x - 3) + C$
An improper fraction: divide first 2 steps
Find ∫x³/(x² + 1) dx.
Show solution
1. Cue: the top’s degree is above the bottom’s. Move: divide.
  
  $x^{3} = x (x^{2} + 1) - x \Rightarrow \frac{x ^{3}}{x ^{2} + 1} = x - \frac{x}{x ^{2} + 1}$
2. Then a power and an f′/f:
  
  $\int (x - \frac{x}{x ^{2} + 1}) d x = \frac{x ^{2}}{2} - \frac{1}{2} ln (x^{2} + 1) + C$
Answer

x²/2 − ½ ln(x² + 1) + C. Two moves, the second a substitution, and neither is visible until the first is made.

$\int \frac{x ^{3}}{x ^{2} + 1} d x = \frac{x ^{2}}{2} - \frac{1}{2} ln (x^{2} + 1) + C$
An area with no cue in it 3 steps
Find ∫₀^π²/4 sin √x dx. Press No cue at all: the page finds the route to the antiderivative in two moves, substitution then parts, and stops there.

Open this problem in the tool
Show solution
1. Substitute u = √x, so x = u² and dx = 2u du. The limits move with it: x = 0 gives u = 0, and x = π²/4 gives u = π/2.
  
  $\int_{0}^{π^{2} /4} sin x d x = \int_{0}^{π /2} 2 u sin u d u$
2. Parts, differentiating 2u and integrating sin u:
  
  $\int 2 u sin u d u = - 2 u cos u + \int 2 cos u d u = 2 sin u - 2 u cos u$
3. Between the new limits, where cos(π/2) = 0 and sin(π/2) = 1:
  
  $[2 sin u - 2 u cos u]_{0}^{π /2} = (2 - 0) - (0 - 0) = 2$
Answer

2, exactly. Both limits carry a π and the answer carries none: the substitution turned π²/4 into π/2, where the sine is 1 and the cosine is 0. Simpson's rule on the original integrand, with 4,000 panels, gives 1.999999 and creeps up on 2 slowly, because √x has an infinite slope at 0.

$\int_{0}^{π^{2} /4} sin x d x = 2$

When the substitution cue and the parts cue both fire, believe substitution 🖖

Press Both cues fire. 2x·e^(x²) is a power times an exponential, the shape usually handed to parts, and parts does reach the table, in three moves: u = e^(x²) with dv = 2x dx leaves 2x³e^(x²), longer than what it started from. Substitution needs one, because x² sits inside the exponential and its derivative 2x is already standing beside it. The page ran the same comparison over 1,100 generated problems. Both cues fired on 113 of them; substitution was the shortest first move on all 113 and parts on none, and on 26 parts never reached the table within three moves. On its own, without the substitution cue, the parts cue was right on 181 of 188.

Most of these integrals can be started more than one way 🖖

Press Two ways in. sin x·cos x falls to substitution, u = sin x, in one move. It falls to parts in one move as well: u = sin x with dv = cos x dx gives the integral back as itself, I = sin²x − I, so I = sin²x/2. And to the trigonometric move, which keeps one factor for du, also in one. Over the 1,100 problems, 842 could be started by two or more of the five moves and still reach the table within three, and on 650 a second route was at most one move longer than the best. Some of those second routes are detours a teacher would not take (parts on a rational function, for one), and they count, because they arrive. So the page never asks for the technique: any first move that gets there is marked right, and the line under the buttons says whether a shorter one exists. Two routes can end in answers that look different; the first page of this chapter, Antiderivatives, shows why both are right.

Length says almost nothing about how many moves an integral needs 🖖

Press Short is not easy. arctan x is two symbols long and takes two moves: parts with dv = 1·dx, which leaves x/(x² + 1), then a substitution. Among the look-alikes, −6x⁴ − 2/√(1 − x²) is fifteen symbols long and is in the table as it stands. Across the 1,100 problems, the ones the table does at once have a median length of 8 symbols, the one-move problems 10, the two-move problems 9; the longest table integral has 15 symbols and the shortest two-move one has 3. What the cues read is arrangement: what sits inside what, which factor gets simpler when differentiated, whether a denominator splits, whether a power of a sine is odd or even.

Learning path

Techniques of integration

References (2)

A strategy for picking the technique, with the page's three findings in it: look for a simple substitution before the heavier techniques (u = x² − 1 for x/(x² − 1), which partial fractions would also do), “a polynomial times a trig function, exponential, or logarithm” as the sign that parts may work, “some integrals can be done in more than one way”, and a substitution that leads on to parts or partial fractions: P. Dawkins, “Integration Strategy”, Paul's Online Notes, Calculus II, section 7.7, Lamar University (page last modified 31 July 2026).
The both-cues integral done by parts: ∫t³e^(t²) dt, where the strict LIATE choice fails and dv = t·e^(t²) dt is taken instead, an integral that is itself a substitution (Example 3.4, “When LIATE Doesn't Quite Work”); and ∫sin(ln x) dx with dv = 1·dx, solved after two rounds (Example 3.5): G. Strang and E. Herman, Calculus Volume 2, OpenStax (2016), section 3.1, “Integration by Parts”.

By the 42Math team · Reviewed against published sources. Corrections welcome.