u-substitution and integration by parts: the one question that chooses between them, the LIATE order for u, and two examples worked line by line

Two techniques cover most first-year integrals. u-substitution undoes the chain rule: if one factor is the derivative of what sits inside another factor, substitute for the inside. Integration by parts undoes the product rule: if the integrand is a product of two unrelated kinds of function, split it as u dv and use ∫ u dv = uv − ∫ v du. One question decides between them: is a factor the derivative of an inside? Here is each technique worked once, with the marks students drop.

The one question that chooses between u-substitution and integration by parts

Look at the integrand and ask: is one factor the derivative of something inside another factor, give or take a constant?

• If yes, use u-substitution, with u as that inside.
• If no, and the integrand is a product of two different kinds of function, use integration by parts.

Here is the question applied to four integrands.

Integrand	Is a factor the derivative of an inside?	Technique
x e^(x²)	Yes: x is half the derivative of x²	Substitution, u = x²
x eˣ	No: the exponent x has derivative 1, and no inside pairs with x	Parts
x ln x	No: the derivative of ln x is 1/x, which is not a factor	Parts
(ln x)/x	Yes: 1/x is the derivative of ln x	Substitution, u = ln x

The two techniques are the chain rule and the product rule run backwards. Substitution undoes the chain rule, which is why it looks for an inside and its derivative. Parts undoes the product rule, which is why it splits a product.

What an antiderivative is, in one paragraph

An antiderivative of f is any function F with F' = f. There is never just one. If x²/2 is an antiderivative of x, so is x²/2 + 5, and so is x²/2 − 100, because a constant differentiates to 0. Every antiderivative of f differs from every other by a constant, so the whole family is written F(x) + C. That fact is a consequence of the Mean Value Theorem, where the + C comes from. The "+ C" on an indefinite integral is the first mark students drop, and it is the cheapest one to keep.

Worked example 1: u-substitution, line by line

Find ∫ x (x² + 1)³ dx.

• Line 1, ask the question. The inside is x² + 1. Its derivative is 2x. The other factor is x, which is half of 2x. A factor is the derivative of an inside, up to a constant, so substitution applies.
• Line 2, write u and du in full. u = x² + 1, so du = 2x dx. Solve for the piece you have: x dx = du/2.
• Line 3, rewrite the whole integral in u. ∫ u³ · (du/2) = (1/2) ∫ u³ du. No x is left.
• Line 4, integrate. (1/2) · u⁴/4 = u⁴/8.
• Line 5, return to x and add the constant. (x² + 1)⁴/8 + C.

The check. Differentiate the answer with the chain rule: 4(x² + 1)³ · 2x / 8 = x(x² + 1)³. That is the integrand, so the answer is right.

If you expand everything and integrate term by term instead, you get x⁸/8 + x⁶/2 + 3x⁴/4 + x²/2 + C. It looks different, but it differs from (x² + 1)⁴/8 only by the constant 1/8, which the C absorbs. Both are correct. This is the "+ C" doing its job.

The common wrong lines

(x² + 1)⁴/4 + C, the du scaling forgotten. Here x dx was treated as if it were du. But du = 2x dx, so x dx is half of du. The check catches it at once: differentiating (x² + 1)⁴/4 gives 2x(x² + 1)³, twice the integrand. Writing du in full on line 2, then solving for the piece you have, prevents this.

Leaving an x in the integral after substituting. If the rewritten integral still contains x, it is neither in x nor in u, and treating that x as a constant gives nonsense. Every x must go, either into u or into du. If you cannot remove them all, substitution was the wrong choice, or u was.

Worked example 2: integration by parts, line by line

Find ∫ x ln x dx.

• Line 1, ask the question. The derivative of ln x is 1/x, and 1/x is not a factor. No factor is the derivative of an inside. The integrand is a product of an algebraic function and a logarithm, so integration by parts applies.
• Line 2, choose u by LIATE. LIATE is the order Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential, and the factor whose type comes earliest becomes u. The logarithm comes before the algebraic x, so u = ln x and dv = x dx.
• Line 3, find du and v. du = (1/x) dx. v = x²/2.
• Line 4, apply the integration by parts formula. ∫ u dv = uv − ∫ v du = (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − ∫ (x/2) dx.
• Line 5, finish the easier integral. (x²/2) ln x − x²/4 + C.

The check. Differentiate, using the product rule on the first term: x ln x + (x²/2)(1/x) − x/2 = x ln x + x/2 − x/2 = x ln x. That is the integrand.

The common wrong lines

u = x and dv = ln x dx. Then you need v = ∫ ln x dx before you can even start, which is a harder integral than the one you were given. LIATE puts the logarithm first for exactly this reason: logs are easy to differentiate and awkward to integrate.

Dropping the minus: (x²/2) ln x + x²/4. The formula is uv minus ∫ v du. With a plus, the check leaves an extra x on the end, so the error shows itself if you run the check.

The one students search for

∫ x eˣ dx, with u = x and dv = eˣ dx, so du = dx and v = eˣ: x eˣ − ∫ eˣ dx = x eˣ − eˣ + C. As a definite integral from 0 to 1, it is (1 · e − e) − (0 − 1) = 0 + 1 = 1.

Siblings to try cold

Cover the worked examples. For each, ask the question first, then work it, then check by differentiating.

• ∫ x³ (x⁴ + 2)⁵ dx
• ∫ x cos x dx
• ∫ x² ln x dx

Check

• Substitution with u = x⁴ + 2, so x³ dx = du/4. The answer is (x⁴ + 2)⁶/24 + C.
• Parts with u = x and dv = cos x dx. The answer is x sin x + cos x + C.
• Parts with u = ln x and dv = x² dx. The answer is (x³/3) ln x − x³/9 + C.

A wrong constant on the first is the du scaling. A wrong sign on the second or third is the dropped minus. Whichever you missed is tonight's card.

LIATE is a guide, not a law: how to choose u

LIATE works because of what happens to each type. The early types (logarithms, inverse trig, powers of x) get simpler when you differentiate them, so they make a good u. The late types (trig functions, exponentials) are easy to integrate and do not get worse, so they make a good dv. Most textbook integrals follow that pattern, and the order picks the right u.

It is not a theorem. Some integrals need a different choice, and some need more than one round. ∫ eˣ sin x dx, for example, needs parts twice and then an equation to solve for the original integral; that method, and trigonometric substitution, come later in the course.

Definite integrals with substitution: change the limits

With a definite integral there are two correct routes after a substitution. Either change the limits to u-values (if u = x² + 1 and x runs from 0 to 1, then u runs from 1 to 2) and evaluate in u, or substitute back to x first and use the original limits. Mixing the two, putting u-limits into an expression in x, is a lost mark that no check by differentiation will catch, so decide which route you are on before you write the limits.

Where this sits in the course, and where it goes next

In a Calculus I course these are idea 22, Integration by substitution, and idea 23, Integration by parts. Before them come idea 19, Antiderivatives and the indefinite integral, and idea 21, The Fundamental Theorem of Calculus; after them comes idea 24, Area between curves. MCV4U has no integration, so in Ontario these arrive in first year: first-year calculus says what changes. Both techniques are also tested in AP Calculus.

Area between curves and idea 25, Volumes of solids of revolution, are these same integrals with limits on them. The Fundamental Theorem explains why evaluating an antiderivative at two limits gives an area at all.

The skill this page teaches is choosing, and choosing is only practised when the problems are mixed. Rohrer and Taylor (2007) found that students who practised shuffled problem types scored higher on a later test than students who practised in blocks. A set of ten substitution problems never asks the question at the top of this page. Mix the problem types: put substitution and parts in one sitting. The worked-example-then-sibling order here follows Sweller and Cooper (1985), and worked examples, then practice problems explains it.

u-substitution and integration by parts on one screen

• The one question: is a factor the derivative of an inside? Yes means substitution; a product of unrelated kinds means parts.
• Substitution: write u and du in full, solve for the piece you have, leave no x behind.
• Parts formula: ∫ u dv = uv − ∫ v du.
• LIATE: Logarithmic, Inverse trig, Algebraic, Trig, Exponential; the earliest becomes u. A guide, not a law.
• Check 1: differentiate your answer and get the integrand back exactly.
• Check 2: count the du scaling; x dx is du/2 when du = 2x dx.
• Always: + C on an indefinite integral; on a definite integral, one route for the limits.