Where does the integration-by-parts formula come from?

From the product rule. Integrating both sides of (uv)′ = u′v + uv′ gives uv = ∫u′v dx + ∫uv′ dx, and moving one integral to the other side leaves ∫uv′ dx = uv − ∫u′v dx. Nothing else is hiding in the formula — it is the product rule read backwards.

What should I take as u when integrating by parts?

Whatever gets simpler when differentiated, provided the other factor can be integrated. In ∫x·eˣ dx take u = x, because its derivative is 1 and eˣ integrates into itself. In ∫ln x dx take u = ln x and the invisible 1 as v′ — differentiating turns the logarithm into 1/x and the integral becomes elementary.

What should I substitute?

The inner function of a composition, provided its derivative stands in the integral as a factor — a constant multiple is fine. In ∫2x·cos(x²) dx substitute u = x², because du = 2x dx and that factor is already there. If the derivative of the inner function is nowhere to be seen, substitution will not help and another route is needed.

Do I have to go back to x in a definite integral?

Not if you change the bounds. The substitution u = g(x) moves the limits from a and b to g(a) and g(b), so the computation finishes in u and nothing has to be undone. You only go back to x when the bounds were left alone — and mixing the two routes is the commonest mistake in this topic.

Why does a linear substitution divide by a?

Because the chain rule multiplies by a, and integration has to undo that. The derivative of F(ax + b) is a·f(ax + b), so to get f(ax + b) alone the antiderivative has to be divided by a. Hence ∫sin(3x) dx = −cos(3x)/3, not −cos(3x).

When is integration by parts applied twice?

When a product is still standing after the first step — in ∫x²eˣ dx, for instance, where the first step lowers the square to a first power and only the second removes it. A polynomial of degree n against an exponential or a trigonometric function needs exactly n steps, because each one drops the degree by one.

How do I tell which of the two techniques to use?

By what the integrand shows. A composition whose inner derivative stands beside it as a factor means substitution. A product of functions of different kinds — a polynomial times an exponential, a polynomial times a sine, a lone logarithm — means parts. When both fit, take the one that leaves the shorter integral on the right.

Can every integral be done with one of these two techniques?

No. They widen the table but do not close it: ∫e^(−x²) dx cannot be written with any finite formula built from elementary functions, even though the function is smooth and integrable. What is left then is a numerical estimate — or, as with the error function, giving that integral a name of its own.

Advanced level

Integration by parts and substitution

The table of integrals stops at polynomials, sine and cosine — and ∫ x·eˣ dx already has no row in it. Two techniques close the gap, because each of them is a differentiation rule read backwards: the product rule gives integration by parts, and the chain rule gives integration by substitution.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Where this is used

Real situations where you count exactly the way this lesson teaches:

Work done stretching a resistance band
The force needed to stretch a rubber band is not constant — it builds up and levels off: F(x) = 150(1 − e^(−4x)) newtons. Work is the integral of force along the distance, so stretching it by 0.5 m takes 150·(0.5 + 0.25·e^(−2) − 0.25) ≈ 42.6 joules. The exponential term has no row in the table of integrals — the linear substitution u = −4x is what handles it.
Energy drawn by a machine on start-up
An electrician records the instantaneous power of a start-up as P(t) = 40·t·e^(−t/2) kilowatts. The energy is the area under that curve, and a variable times an exponential has no row in the table: integration by parts gives the antiderivative −2t·e^(−t/2) − 4e^(−t/2), so over the first 4 hours the machine eats 40·(4 − 12·e^(−2)) ≈ 95 kilowatt-hours.
How much water leaves the tank
Flow through an opening falls off with the pressure behind it: q(t) = 30·e^(−t/10) litres per minute. The volume that leaves over 20 minutes is the integral of that flow rate, namely 300·(1 − e^(−2)) ≈ 259 litres. The substitution u = −t/10 turns it into the exponential integral you already know.
A patient exposure to a drug
A pharmacologist computes the area under the concentration curve for the model C(t) = 8t·e^(−0.5t) milligrams per litre. Over the first 6 hours, integration by parts gives 32·(1 − 4·e^(−3)) ≈ 25.6 mg·h/l — the number compared against the exposure of a reference dose. The same variable-times-exponential product, the same technique.

All formulas

Integration by parts
$\int u(x)\,v'(x)\,dx = u(x)\,v(x) - \int u'(x)\,v(x)\,dx$
the product rule, reversed
Parts, definite
$\int_a^b u\,v'\,dx = \Big[u\,v\Big]_a^b - \int_a^b u'\,v\,dx$
the uv term is evaluated at the bounds, the rest stays an integral
Integration by substitution
$\int f\big(g(x)\big)\,g'(x)\,dx = F\big(g(x)\big) + C$
the chain rule, reversed
The bounds travel
$\int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$
a substitution moves the limits too
Linear substitution
$\int f(ax+b)\,dx = \frac{1}{a}\,F(ax+b) + C$
the commonest case: divide by a
The exponential integral
$\int e^x\,dx = e^x + C$
needed by the worked examples on parts

The table of integrals in the lesson on the integral stops at polynomials, sine and cosine. Yet $\int x e^x\,dx$ and $\int (2x+1)^5\,dx$ have no row in it — and both are integrals of functions you know how to differentiate. That is where both techniques in this lesson come from: each of them is one differentiation rule read backwards.

differentiation rule	integration technique
product rule $(uv)' = u'v + uv'$	integration by parts
chain rule $\big(F(g(x))\big)' = f(g(x))\,g'(x)$	integration by substitution

One antiderivative from outside the table will be needed along the way, straight from the lesson on exponential and logarithmic functions: since $(e^x)' = e^x$ ,

\int e^x\,dx = e^x + C

Integration by substitution

The chain rule says that the derivative of a composition $F(g(x))$ is $f(g(x)) \cdot g'(x)$ . Read that equality from right to left:

\int f\big(g(x)\big)\,g'(x)\,dx = F\big(g(x)\big) + C

In practice the inner function is given a name. Substitute $u = g(x)$ , so that $du = g'(x)\,dx$ — and the whole integral moves over to $u$ :

\int f(u)\,du = F(u) + C

Evaluate ∫ 2x·cos(x²) dx.

What to substitute

There is one rule: substitute the inner function of a composition, provided its derivative stands in the integral as a factor. A constant multiple is fine — the missing factor moves out in front of the integral.

integral	substitute	why
$\int x\sqrt{x^2+1}\,dx$	$u = x^2+1$	$du = 2x\,dx$ , and $x\,dx$ is there
$\int \dfrac{\ln x}{x}\,dx$	$u = \ln x$	$du = \tfrac{1}{x}\,dx$ , and $\tfrac{1}{x}$ is there
$\int \sin^3 x \cos x\,dx$	$u = \sin x$	$du = \cos x\,dx$ , and $\cos x$ is there
$\int (2x+1)^5\,dx$	$u = 2x+1$	$du = 2\,dx$ , a bare constant

The last row is so common that it has a shortcut of its own. When the inside is a linear expression:

\int f(ax+b)\,dx = \frac{1}{a}\,F(ax+b) + C

Dividing by $a$ is unavoidable: the chain rule multiplies by $a$ , so integration has to take it back. Hence $\int \sin 3x\,dx = -\tfrac{1}{3}\cos 3x + C$ , not $-\cos 3x$ .

A definite integral: the bounds are substituted too

In a definite integral a substitution changes not only the function but the limits of integration as well — because the variable being integrated over is now a different one:

\int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du

Evaluate ∫₀¹ (2x + 1)² dx by changing the limits of integration.

The drawings show what actually happened: the region changed shape, not size.

Before the substitution: a narrow, tall region under y = (2x + 1)² between x = 0 and x = 1.

After the substitution: the region has stretched sideways and flattened, because the bounds moved to 1 and 3 and the function to u²/2. Both regions have area 13/3.

Integration by parts

The second technique reverses the product rule. Start from it and integrate both sides:

\big(u\,v\big)' = u'v + u\,v' \quad \Longrightarrow \quad u\,v = \int u'v\,dx + \int u\,v'\,dx

Move one integral across and the formula is there:

\int u(x)\,v'(x)\,dx = u(x)\,v(x) - \int u'(x)\,v(x)\,dx

The formula does not evaluate the integral — it trades it for another one. All the effort goes into making that second one easier than the first.

Choosing u

Take as $u$ the factor that gets simpler when differentiated; as $v'$ , the one you know how to integrate.

integrand	$u$	$v'$	what it buys
$x\,e^x$	$x$	$e^x$	$u' = 1$ , the product disappears
$x\sin x$	$x$	$\sin x$	as above
$\ln x$	$\ln x$	$1$	the logarithm turns into $\tfrac{1}{x}$
$x^2 e^x$	$x^2$	$e^x$	the degree drops by one per step

Evaluate ∫ x·eˣ dx.

Evaluate ∫ ln x dx.

When it takes two rounds

If a product is still standing after the first step, the technique is applied again. A polynomial of degree $n$ against an exponential takes exactly $n$ steps — each one drops the degree by one.

Evaluate ∫ x²·eˣ dx.

The formula as a picture

For an increasing function that starts at zero, the parts formula can be drawn. Take $y = x^2$ on $\langle 0;\, 1 \rangle$ together with its inverse $y = \sqrt{x}$ :

The curve y = x² cuts the unit square in two. The shaded part lies below it; the other one, above it, is at the same time the region under the inverse curve y = √x.

The area under the parabola is $\int_0^1 x^2\,dx = \tfrac{1}{3}$ , the area above it is $\int_0^1 \sqrt{x}\,dx = \tfrac{2}{3}$ . Together they make $1$ , the area of the whole square — and that is the integration-by-parts formula written without a single letter:

\int_0^1 x^2\,dx + \int_0^1 \sqrt{x}\,dx = 1 \cdot 1

Which technique to reach for

See a composition with the derivative of its inside next to it? Substitution.
See a product of functions of different kinds — a polynomial times an exponential, a polynomial times a sine, a lone logarithm? Parts.
See both at once? Take the one that leaves the shorter integral on the right; if one step made things worse, go back and try the other.
See neither? Try reshaping the integrand first — split the fraction, use a trigonometric identity, pull a constant out in front.

It is worth remembering what all this is computed for: the work done by a variable force is the integral $W = \int F(x)\,dx$ , and its result comes out in joules — a unit the energy converter will convert for you.

Exercises

The set mixes three skills: a linear substitution (the answer is graded up to the constant, so the $+\,C$ may be left off), integration by parts across three families ( $x e^x$ , $x\sin x$ , $\ln x$ ), and a definite integral evaluated by substitution, where the answer is a number or a fraction. An indefinite integral is typed as a formula, e.g. (2x+1)^4/8 or x*e^x-e^x.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

∫ (2x + 1)³ dx =

What the set deliberately does not ask is the choice of technique together with its justification: the answer to that is a sentence, not a number. You practise it by reading the four points above and checking, on each integral, which one applies.

Common mistakes

Forgetting to move the bounds in a definite integral — after a substitution the limits have to be values of $u$ , not of $x$ ; otherwise you are measuring the area over a different interval entirely.
Substituting with no derivative inside — if $g'(x)$ does not stand in the integral as a factor, the substitution leaves two variables in the expression at once and cannot be finished.
Skipping the division by $a$ in a linear substitution — $\int\cos 5x\,dx$ is $\tfrac{1}{5}\sin 5x + C$ , not $\sin 5x + C$ .
Choosing $u$ the wrong way round — if the integral grew after the first step, the roles were reversed; go back and swap the factors.
Losing the minus in the parts formula — a minus follows the $uv$ term, and with $v = -\cos x$ the two minus signs multiply, which is where a sign error is easiest to make.
Writing the constant mid-computation — with integration by parts, $+\,C$ is written once, at the very end.

Formula card

Topic: Integration by parts and substitution

Integration by parts
$\int u(x)\,v'(x)\,dx = u(x)\,v(x) - \int u'(x)\,v(x)\,dx$
the product rule, reversed
Parts, definite
$\int_a^b u\,v'\,dx = \Big[u\,v\Big]_a^b - \int_a^b u'\,v\,dx$
the uv term is evaluated at the bounds, the rest stays an integral
Integration by substitution
$\int f\big(g(x)\big)\,g'(x)\,dx = F\big(g(x)\big) + C$
the chain rule, reversed
The bounds travel
$\int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$
a substitution moves the limits too
Linear substitution
$\int f(ax+b)\,dx = \frac{1}{a}\,F(ax+b) + C$
the commonest case: divide by a
The exponential integral
$\int e^x\,dx = e^x + C$
needed by the worked examples on parts

The region under y = (2x + 1)² between 0 and 1. The substitution u = 2x + 1 turns it into the region on the next drawing — the shape changes, the area does not.

The same region after the substitution u = 2x + 1. The bounds moved from 0 and 1 to 1 and 3, and the integrand flattened into u²/2. Both areas equal 13/3.

The unit square cut in two by y = x². The area below it is 1/3, the area above it — which is the area under the inverse curve y = √x — is 2/3, and together they make the whole square. This is the integration-by-parts formula drawn rather than written.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min