Mathematics · Calculus / Integration

Integration by Parts

Q: How do I integrate ln x with integration by parts?

With the 1 trick: write ln x as the product $1\cdot{}\ln{} x$. Choose u = ln x (easy to differentiate) and v′ = 1 (easy to integrate), so u′ = 1/x and $v = x$. The formula gives $\int{}\ln{} x dx = x\cdot{}\ln{} x - \int{}x\cdot{}(\frac{1}{x}) dx = x\cdot{}\ln{} x - \int{}1 dx = x\cdot{}\ln{} x - x + C$. Check: $(x\cdot{}\ln{} x - x)$′ = ln x + x·(1/x) − 1 = ln x ✓. The trick works whenever a function is hard to integrate but easy to differentiate; the antiderivatives of arctan x or arcsin x are found the same way. In exams ∫ln x dx is a classic, often part of an area task with the ln curve, and without the 1 trick you cannot even start.

Q: What if a product remains after one round of integration by parts?

Then apply the method again. For ∫x²·eˣ dx the first round $(u = x^{2})$ lowers the degree to ∫2x·eˣ dx, the second $(u = 2x)$ to ∫2·eˣ dx, which is elementary. Result: (x² − 2x + 2)eˣ + C. It is important to keep the roles consistent: always differentiate the polynomial, otherwise you undo the first step. A special case is the "phoenix": for ∫eˣ·sin x dx the original integral I itself reappears after two rounds, with factor −1. You solve the resulting equation I = eˣ(sin x − cos x) − I algebraically: 2I = eˣ(sin x − cos x), so I = ½eˣ(sin x − cos x) + C.

Q: Where does the integration by parts formula come from?

It is the integrated product rule. The product rule says $(u\cdot{}v)$′ = u′·v + u·v′. Integrating both sides over x, the left side becomes u·v (integration undoes differentiation) and the right side the sum of the two integrals: $u\cdot{}v = \int{}u$′·v dx + ∫u·v′ dx. Solving for the second integral produces the familiar formula $\int{}u\cdot{}v$′ $dx = u\cdot{}v - \int{}u$′·v dx. This understanding is more than theory: it explains why the minus sign is there (it comes from rearranging), why an "integrated-out" boundary term $u\cdot{}v$ appears, and why the method is the counterpart of substitution, which in turn reverses the chain rule. For definite integrals the boundary term becomes [u·v]ₐᵇ evaluated at the limits.

Integration by parts integrates products by reversing the product rule of differentiation.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Formula

\int u \cdot v^{'} d x = u \cdot v - \int u^{'} \cdot v d x

Dimensionless (calculus)

Variables & units

Variables & units
Symbol	Meaning	Unit
$u$	Factor that is differentiated (becomes simpler)	dimensionless
v'	Factor that is integrated	dimensionless
$v$	Antiderivative of v′	dimensionless

LaTeX

\int u \cdot v' \, dx = u \cdot v - \int u' \cdot v \, dx

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Derivation & background

Integrating the product rule $(uv)' = u'v + uv'$ and rearranging yields integration by parts. Strategy: choose u so that u′ becomes simpler (polynomials), and v′ so that it stays easy to integrate (e^x, sin, cos). A classic: $\int \ln x\,dx$ via the trick $1\cdot{}\ln{} x$ with $u = \ln{} x$ . Sometimes two applications are needed, for example for $\int{}x^{2}\cdot{}e^{x} dx$ or the circular trick for $\int{}e^{x}\cdot{}\sin{} x dx$ .

Exam blueprint

Validity range

Valid when u and v are continuously differentiable on the interval. Useful only if the new integral ∫u′v dx is simpler than the original one.

Derivation steps

Integrate the product rule and solve for ∫uv′ dx.

1 $(u\cdot{}v)$ ′ = u′v + uv′; integrate both sides: $u\cdot{}v = \int{}u$ ′v dx + ∫uv′ dx.
2Rearranging gives ∫uv′ $dx = u\cdot{}v - \int{}u$ ′v dx.

Rearrangements

Definite integral

\int_{a}^{b} u v' \, dx = [u v]_{a}^{b} - \int_{a}^{b} u' v \, dx

The integrated-out part is evaluated at the limits.

ln trick

\int \ln x \, dx = x \ln x - x + C

With $u = \ln{} x$ and v′ = 1; a standard exam task.

Task variant

Evaluate $\int{}x\cdot{}\sin{} x dx$ .

$u = x$ , v′ = sin x, $v = -\cos{} x$ : $\int{}x\cdot{}\sin{} x dx = -x\cdot{}\cos{} x + \int{}\cos{} x dx = -x\cdot{}\cos{} x + \sin{} x + C$ .
Check by differentiating: $x\cdot{}\sin{} x$ ✓.

Evaluate ∫₀¹ x·eˣ dx.

Antiderivative (x − 1)eˣ. [(x − 1)eˣ]₀¹ = 0·e − (−1)·1 = 0 + 1 = 1.

Common mistakes

Choosing u and v′ badly, e.g. u = eˣ for ∫x·eˣ dx.

Let polynomials be differentiated $(u = x)$ , integrate exponential/sin/cos.

Forgetting the minus sign in front of the remaining integral.

Formula: $u\cdot{}v MINUS \int{}u$ ′v dx.

Determining v wrongly: v′ = sin x gives $v = -\cos{} x$ , not cos x.

Check v by differentiating: (−cos x)′ = sin x ✓.

Giving up after one application.

For x²·eˣ apply twice; for eˣ·sin x return to the original integral after two rounds and solve for it.

Exam context

Advanced-course standard: integrals with x·eˣ, $x\cdot{}\sin{} x$ or ln x, often combined with area tasks.

Formula cluster

Integration techniques

Reverses the product rule, as substitution reverses the chain rule.

(fg)' = f'g + fg'

Product Rule of Differentiation

∫xⁿ dx = xⁿ⁺¹/(n+1) + C

Antiderivatives and Basic Integrals

Worked example

$\int xe^x\,dx$ : $u=x$ , $v'=e^x$ . Result: $xe^x-\int e^x\,dx=xe^x-e^x+C=(x-1)e^x+C$ . Check: $((x-1)e^x)'=e^x+(x-1)e^x=xe^x$ ✓.

Applications

Integrals of products (x·e^x, $x\cdot{}\sin{} x$ , ln x), expected values of continuous distributions, Fourier coefficients, physics (centroids)

How do you calculate with Integration by Parts?

Here is how to work through a typical Integration by Parts (∫u·v' dx = u·v − ∫u'·v dx) task step by step:

1
Task
Evaluate $\int{}x\cdot{}\sin{} x dx$ .
Solution path
1. $u = x$ , v′ = sin x, $v = -\cos{} x$ : $\int{}x\cdot{}\sin{} x dx = -x\cdot{}\cos{} x + \int{}\cos{} x dx = -x\cdot{}\cos{} x + \sin{} x + C$ .
2. Check by differentiating: $x\cdot{}\sin{} x$ ✓.
2
Task
Evaluate ∫₀¹ x·eˣ dx.
Solution path
Antiderivative (x − 1)eˣ. [(x − 1)eˣ]₀¹ = 0·e − (−1)·1 = 0 + 1 = 1.

Frequently asked questions about Integration by Parts

How does integration by parts work step by step?+

Split the integrand into two factors u and v′. Differentiate u (giving u′) and integrate v′ (giving v). Then insert into the formula $\int{}u\cdot{}v$ ′ $dx = u\cdot{}v - \int{}u$ ′·v dx and solve the new, hopefully simpler integral. Example ∫x·eˣ dx: choose $u = x$ , v′ = eˣ, so u′ = 1, v = eˣ. Substituting: x·eˣ − ∫1·eˣ dx = x·eˣ − eˣ + C = (x − 1)eˣ + C. Check by differentiating: eˣ + (x − 1)eˣ = x·eˣ ✓. The method is called "by parts" because only part of the product is integrated; the rest is shifted into a new integral, which must be easier to solve, otherwise the split was badly chosen.

How do I choose u and v′ correctly?+

Guiding idea: u should become simpler when differentiated, v′ must be easy to integrate. Polynomials are ideal u candidates, since their degree drops with every differentiation: x becomes 1, and the remaining integral becomes trivial. Exponential, sine and cosine are ideal v′ candidates, because integrating them does not make them more complicated. So for $\int{}x\cdot{}\sin{} x dx$ choose $u = x$ , v′ = sin x. Logarithms are the exception: ln x is hard to integrate directly but easy to differentiate, so ln x always becomes u, if necessary with the trick $\int{}\ln{} x dx = \int{}1\cdot{}\ln{} x dx$ . The priority list LIATE (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) serves as a mnemonic: whatever stands further left becomes u.

How do I integrate ln x with integration by parts?+

With the 1 trick: write ln x as the product $1\cdot{}\ln{} x$ . Choose u = ln x (easy to differentiate) and v′ = 1 (easy to integrate), so u′ = 1/x and $v = x$ . The formula gives $\int{}\ln{} x dx = x\cdot{}\ln{} x - \int{}x\cdot{}(\frac{1}{x}) dx = x\cdot{}\ln{} x - \int{}1 dx = x\cdot{}\ln{} x - x + C$ . Check: $(x\cdot{}\ln{} x - x)$ ′ = ln x + x·(1/x) − 1 = ln x ✓. The trick works whenever a function is hard to integrate but easy to differentiate; the antiderivatives of arctan x or arcsin x are found the same way. In exams ∫ln x dx is a classic, often part of an area task with the ln curve, and without the 1 trick you cannot even start.

What if a product remains after one round of integration by parts?+

Then apply the method again. For ∫x²·eˣ dx the first round $(u = x^{2})$ lowers the degree to ∫2x·eˣ dx, the second $(u = 2x)$ to ∫2·eˣ dx, which is elementary. Result: (x² − 2x + 2)eˣ + C. It is important to keep the roles consistent: always differentiate the polynomial, otherwise you undo the first step. A special case is the "phoenix": for ∫eˣ·sin x dx the original integral I itself reappears after two rounds, with factor −1. You solve the resulting equation I = eˣ(sin x − cos x) − I algebraically: 2I = eˣ(sin x − cos x), so I = ½eˣ(sin x − cos x) + C.

Where does the integration by parts formula come from?+

It is the integrated product rule. The product rule says $(u\cdot{}v)$ ′ = u′·v + u·v′. Integrating both sides over x, the left side becomes u·v (integration undoes differentiation) and the right side the sum of the two integrals: $u\cdot{}v = \int{}u$ ′·v dx + ∫u·v′ dx. Solving for the second integral produces the familiar formula $\int{}u\cdot{}v$ ′ $dx = u\cdot{}v - \int{}u$ ′·v dx. This understanding is more than theory: it explains why the minus sign is there (it comes from rearranging), why an "integrated-out" boundary term $u\cdot{}v$ appears, and why the method is the counterpart of substitution, which in turn reverses the chain rule. For definite integrals the boundary term becomes [u·v]_aᵇ evaluated at the limits.

Quanta exam set

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Which formula describes Integration by Parts?

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Question (front)

How do you rearrange Integration by Parts for Definite integral?

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Which common mistake happens with Integration by Parts?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Calculus Volume 2, §3.1: Integration by Parts

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\int_{a}^{b} f(x) \, dx = F(b) - F(a)

Fundamental Theorem of Calculus

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Chain Rule of Differentiation

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n

Taylor Series

\frac{d}{dx}(x^n) = n \cdot x^{n-1}

Power Rule of Differentiation

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Symbol

Meaning

Unit

u

Factor that is differentiated (becomes simpler)

dimensionless

Factor that is integrated

dimensionless

v

Antiderivative of v′

dimensionless

Derivation & background

Exam blueprint

Validity range

Valid when u and v are continuously differentiable on the interval. Useful only if the new integral ∫u′v dx is simpler than the original one.

Derivation steps

Integrate the product rule and solve for ∫uv′ dx.

1 $(u\cdot{}v)$ ′ = u′v + uv′; integrate both sides: $u\cdot{}v = \int{}u$ ′v dx + ∫uv′ dx.
2Rearranging gives ∫uv′ $dx = u\cdot{}v - \int{}u$ ′v dx.

How do you calculate with Integration by Parts?

Here is how to work through a typical Integration by Parts (∫u·v' dx = u·v − ∫u'·v dx) task step by step:

Task

Evaluate $\int{}x\cdot{}\sin{} x dx$ .

Solution path

$u = x$ , v′ = sin x, $v = -\cos{} x$ : $\int{}x\cdot{}\sin{} x dx = -x\cdot{}\cos{} x + \int{}\cos{} x dx = -x\cdot{}\cos{} x + \sin{} x + C$ .
Check by differentiating: $x\cdot{}\sin{} x$ ✓.

Task

Evaluate ∫₀¹ x·eˣ dx.

Solution path

Antiderivative (x − 1)eˣ. [(x − 1)eˣ]₀¹ = 0·e − (−1)·1 = 0 + 1 = 1.

Frequently asked questions about Integration by Parts

How does integration by parts work step by step?+

How do I choose u and v′ correctly?+

How do I integrate ln x with integration by parts?+

What if a product remains after one round of integration by parts?+

Where does the integration by parts formula come from?+

Retain Integration by Parts for exams

Create a curated FSRS exam set for ∫u·v' dx = u·v − ∫u'·v dx: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential