In ∫x log x dx, integration by parts should differentiate:

Neither — parts does not apply

Integration and Definite Integrals

Differentiation is a procedure; integration is a search. There is a rule for differentiating any expression you can write down, but no such rule for integrating one, so the skill here is pattern recognition — deciding which of five or six standard techniques a given integral is asking for. That is why practice matters more in this chapter than in any other, and why it is worth learning the signals: a product of unlike functions suggests integration by parts, a composite suggests substitution, a rational function suggests partial fractions. Definite integrals then add a second layer, where symmetry properties can make a hard integral collapse to nothing at all.

Choosing a Technique

What the integrand is telling you
Signal in the integrand	Technique
A composite function with its derivative present	Substitution
A product of two unlike functions	Integration by parts, ∫u dv = uv − ∫v du
A proper rational function	Partial fractions
A quadratic under a root or in a denominator	Complete the square, then a standard form
Only sines and cosines, one of odd power	Substitute for the even-powered one
a sin x + b cos x in a denominator	Substitute t = tan(x/2)

🧠 ILATE fixes which factor to differentiate in integration by parts

Integration by parts works only if the second integral is easier than the first, which means you must differentiate the factor that simplifies and integrate the one that stays manageable. The order Inverse, Logarithmic, Algebraic, Trigonometric, Exponential tells you which to take as u: whichever appears earlier in that list. So in ∫x log x dx the logarithm is differentiated and x integrated, while in ∫x eˣ dx the algebraic x is differentiated. Getting the choice backwards is not merely slower — it usually produces a harder integral than the one you started with, which is a useful signal that you picked wrongly and should swap.

Two standard results are worth having ready because they recur constantly: ∫eˣ(f(x) + f′(x)) dx = eˣf(x), which turns an apparently awkward integral into one line whenever you can spot the pattern, and ∫dx/(x² + a²) = (1/a)tan⁻¹(x/a). Note also that ∫dx/x = log|x| with the modulus, not log x — the domain of the logarithm requires it, and omitting the modulus is a small but frequently penalised slip.

Definite Integrals and Their Properties

Properties that shorten the work
Property	Statement
Reversal	∫ₐᵇ f = −∫_bᵃ f
Splitting	∫ₐᵇ f = ∫ₐᶜ f + ∫_cᵇ f for any c
King property	∫ₐᵇ f(x)dx = ∫ₐᵇ f(a + b − x)dx
Odd function on a symmetric interval	∫_−aᵃ f = 0
Even function on a symmetric interval	∫_−aᵃ f = 2∫₀ᵃ f
Periodic f with period T	∫₀^nT f = n∫₀^T f

🎯 Why the king property turns some impossible integrals into trivial ones

Replacing x by a + b − x reflects the interval onto itself, so the integral is unchanged — but the integrand may look completely different. Adding the original and the reflected version then sometimes cancels the difficult part entirely. The classic case is ∫₀^π/2 dx/(1 + tan x): reflecting gives ∫₀^π/2 dx/(1 + cot x), and the two integrands sum to exactly 1, so twice the integral is π/2 and the answer is π/4 — obtained without ever integrating anything. The signal to try this is an integrand that resists all normal techniques on a symmetric-looking interval such as [0, π/2] or [0, π].

⚠️ Area is not the same as the value of the integral

A definite integral counts area below the axis as negative, so for f(x) = sin x on [0, 2π] the integral is exactly zero while the enclosed area is 4. If a question asks for the area bounded by a curve, you must locate where the curve crosses the axis, integrate over each piece separately and add the absolute values. The same care applies to the area between two curves: integrate (upper − lower), and if the curves cross within the interval, split at the crossing point, because past it the roles reverse. Reporting a signed integral as an area is one of the most reliably penalised errors in this chapter.

Two further tools appear. Leibniz's rule differentiates an integral whose limits contain the variable: d/dx ∫_a(x)^b(x) f(t)dt = f(b(x))b′(x) − f(a(x))a′(x), which is how questions that define a function as an integral are attacked. And a definite integral can be evaluated as the limit of a sum, which is occasionally the intended method when the expression is presented as lim (1/n)Σ f(r/n) — recognising that as ∫₀¹ f(x)dx converts a hard limit into a routine integral.

Key takeaways

Read the integrand for a signal: composite means substitution, product means parts, rational means partial fractions.
ILATE decides which factor to differentiate; a harder second integral means you chose the wrong one.
∫eˣ(f + f′)dx = eˣf is worth spotting, and ∫dx/x = log|x| with the modulus.
The king property ∫ₐᵇf(x)dx = ∫ₐᵇf(a+b−x)dx can cancel the hard part when added to the original.
An odd function integrates to zero over [−a, a]; an even one gives twice the half-interval integral.
A definite integral counts area below the axis as negative — for area, split and take absolute values.
For the area between curves integrate (upper − lower), splitting wherever the curves cross.
Leibniz differentiates an integral with variable limits; lim (1/n)Σf(r/n) is ∫₀¹f(x)dx.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The value of ∫_−π^π sin³x dx is:
1. 4/3
2. 0
3. 2
4. π
Show answer
Answer: B — 0
sin³x is odd, since replacing x by −x changes the sign, and the interval [−π, π] is symmetric about the origin, so the contributions from the two halves cancel exactly and the integral is zero. Spotting oddness first turns this into a one-line answer; integrating sin³x directly is possible but wholly unnecessary. Note that the enclosed area is not zero — only the signed integral is.
In ∫x log x dx, integration by parts should differentiate:
1. x
2. log x
3. Either works equally well
4. Neither — parts does not apply
Show answer
Answer: B — log x
By ILATE the logarithmic factor comes before the algebraic one, so log x is taken as u and differentiated to 1/x while x is integrated to x²/2; the remaining integral is then just ∫x/2 dx. Choosing the other way round would require integrating log x, which is harder than the original problem — and that difficulty is itself the signal that the choice was wrong.
The area enclosed by y = sin x and the x-axis from 0 to 2π is:
1. 0
2. 2
3. 4
4. 2π
Show answer
Answer: C — 4
The integral over [0, 2π] is zero because the second arch lies below the axis and cancels the first, but area is a positive quantity, so you must split at π and add the absolute values: 2 + 2 = 4. This is the standard distinction between the value of a definite integral and the area it is often mistaken for, and the option 0 is offered precisely to catch the substitution of one for the other.
∫₀^π/2 dx/(1 + tan x) equals:
1. π/2
2. π/4
3. 1
4. log 2
Show answer
Answer: B — π/4
Applying the king property with a = 0 and b = π/2 replaces tan x by cot x, and the two integrands add to exactly 1, so 2I = ∫₀^π/2dx = π/2 and I = π/4. No antiderivative is ever needed. The signal to try this was an integrand that resists substitution and parts on an interval symmetric under x → π/2 − x.
∫dx/x equals:
1. log x + C
2. log|x| + C
3. 1/x + C
4. x log x + C
Show answer
Answer: B — log|x| + C
The antiderivative must be valid wherever 1/x is defined, including x < 0, and log x is undefined there — so the modulus is required rather than decorative. Omitting it gives an answer that is correct on the positive half-line only, and questions involving definite integrals over negative ranges are set specifically to expose the omission.
∫eˣ(sin x + cos x) dx equals:
1. eˣ sin x + C
2. eˣ cos x + C
3. eˣ(sin x − cos x) + C
4. 2eˣ sin x + C
Show answer
Answer: A — eˣ sin x + C
The bracket has the form f + f′ with f = sin x, so the standard result ∫eˣ(f + f′)dx = eˣf applies directly and gives eˣ sin x. Recognising the pattern avoids two rounds of integration by parts, which would reach the same answer far more slowly. The same result read backwards is a quick way to differentiate-check any answer of this shape.
If f is periodic with period T, then ∫₀^3T f(x)dx equals:
1. ∫₀^T f(x)dx
2. 3∫₀^T f(x)dx
3. 0
4. T∫₀^T f(x)dx
Show answer
Answer: B — 3∫₀^T f(x)dx
Each successive period contributes exactly the same amount, so integrating over three full periods gives three times the integral over one. This makes long-interval integrals of periodic functions immediate, and it combines usefully with the odd-function property: if the function is also odd over each period, the total is zero.
lim(n→∞) (1/n)[f(1/n) + f(2/n) + … + f(n/n)] equals:
1. f(1)
2. ∫₀¹ f(x)dx
3. ∫₀^n f(x)dx
4. The limit does not exist
Show answer
Answer: B — ∫₀¹ f(x)dx
The expression is a Riemann sum with n equal subintervals of width 1/n on [0, 1], so its limit is the definite integral of f from 0 to 1. Recognising a sum in this shape as an integral is the intended route whenever a limit of a sum appears with 1/n outside and r/n inside — attempting to evaluate the sum directly is almost always hopeless.