Source: University of Texas at Austin, Calc 2 lecture notes
Difficulty: Intermediate | Prerequisites: Calc 1 (product rule, chain rule, basic antiderivatives, Fundamental Theorem of Calculus).
Integration by parts is the integral counterpart of the product rule from differentiation. It is one of the core techniques in Calc 2 for handling integrals that cannot be solved by simple substitution, and it comes up constantly in later courses (differential equations, Fourier analysis, probability). You should already be comfortable differentiating products and computing basic antiderivatives of polynomials, exponentials, trig functions and logarithms. If u-substitution is the first tool you reach for, integration by parts is the second.
Integration by parts rewrites a hard integral as the product of two functions minus a (hopefully easier) integral. You pick one factor to differentiate (u) and one to integrate (dv), then apply the formula ∫ u dv = uv − ∫ v du. The trick is choosing u and dv so the new integral is simpler than the original.
Integration by parts (IBP)
A technique that transforms the integral of a product of two functions into a boundary term minus a new integral. Derived from the product rule for differentiation.
In simple terms: you trade one integral you cannot do for another one you can.
u and dv
The two pieces you split the integrand into. u is the part you will differentiate; dv is the part you will integrate.
Think of it as: u is the piece that gets simpler when you take its derivative, and dv is the piece you can actually antidifferentiate.
LIATE rule (mnemonic)
A priority order for choosing u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Pick u from whichever category comes first in this list.
In simple terms: logs and inverse trig almost always become u, because they simplify nicely when differentiated and are painful to integrate.
Tabular integration (repeated IBP)
A shortcut for integrals where u must be differentiated more than once (e.g. polynomials times exponentials). You build a table of successive derivatives and antiderivatives and alternate signs.
Cycling integrals
When applying IBP twice brings the original integral back on the right-hand side, you can move it to the left and solve algebraically. Classic example: ∫ eˣ sin x dx.
Fundamental Theorem of Calculus (FTC)
Used to evaluate definite integrals after finding the antiderivative via IBP. Compute the antiderivative, then plug in the upper and lower bounds and subtract.
Start from the product rule: d/dx [f(x)g(x)] = f(x)g'(x) + g(x)f'(x). Integrate both sides and rearrange:
∫ f(x)g'(x) dx = f(x)g(x) − ∫ g(x)f'(x) dx
In shorthand notation, with u = f(x) and dv = g'(x) dx:
∫ u dv = uv − ∫ v du
This is the formula you will use on every IBP problem.
The entire game is picking u and dv so that ∫ v du is easier than the integral you started with. The LIATE mnemonic gives a reliable default:
Logarithmic functions (ln x, log x) make good choices for u because their derivatives are algebraic.
Inverse trig functions (arctan x, arcsin x) also simplify on differentiation.
Algebraic functions (x, x², polynomials) lose a degree each time you differentiate, so they eventually vanish.
Trigonometric functions (sin x, cos x) cycle under differentiation and integration, so they work as either u or dv depending on context.
Exponential functions (eˣ, e³ˣ) are easiest to integrate, so they almost always become dv.
∫ 4x e³ˣ dx
Let u = x, dv = e³ˣ dx
Then du = dx, v = e³ˣ/3
Apply IBP: 4[x · e³ˣ/3 − ∫ e³ˣ/3 dx] = 4[x e³ˣ/3 − e³ˣ/9]
Result: (4/3) e³ˣ (x − 1/3) + C
∫ x ln x dx
Let u = ln x (logarithms become u), dv = x dx
Then du = (1/x) dx, v = x²/2
Apply IBP: ln(x) · x²/2 − ∫ (x²/2)(1/x) dx = (x²/2) ln x − ∫ x/2 dx
Result: (x²/2) ln x − x²/4 + C
∫ ln|x| dx
Let u = ln x, dv = 1 dx (set dv = dx when there is no obvious second factor)
Then du = (1/x) dx, v = x
Apply IBP: x ln x − ∫ 1 dx
Result: x ln x − x + C
The strategy: find the antiderivative first (ignore the bounds), then evaluate using the Fundamental Theorem of Calculus.
Example: ∫ from π/2 to π of x cos x dx
u = x, dv = cos x dx, so v = sin x, du = dx
Antiderivative: x sin x + cos x
Evaluate: [π sin π + cos π] − [π/2 sin(π/2) + cos(π/2)] = [0 − 1] − [π/2 + 0] = −1 − π/2
Example: ∫ from 0 to 5 of t e⁻ᵗ dt
u = t, dv = e⁻ᵗ dt, v = −e⁻ᵗ, du = dt
Antiderivative: e⁻ᵗ(−t − 1)
Evaluate: e⁻⁵(−6) − (−1) = 1 − 6/e⁵
∫ (x² + x) eˣ dx
First pass: u = x² + x, dv = eˣ dx → (x² + x)eˣ − ∫ (2x + 1)eˣ dx
Second pass on the new integral: u = 2x + 1, dv = eˣ dx → (2x + 1)eˣ − ∫ 2eˣ dx
Combine: (x² + x)eˣ − (2x + 1)eˣ + 2eˣ = eˣ(x² + x − 2x − 1 + 2) − wait, correcting: eˣ(x² + x − 2x − 1 + 2) simplifies. From the notes the result is:
Result: eˣ(x² − x − 1) + C
When u is a polynomial, you will need to apply IBP as many times as the degree of the polynomial. Each pass reduces the degree by one until only the exponential (or trig) factor remains.
∫ eˣ sin x dx
Some integrals do not simplify on repeated IBP. Instead, the original integral reappears on the right-hand side. When that happens, collect like terms and solve algebraically.
First pass: u = sin x, dv = eˣ dx → eˣ sin x − ∫ eˣ cos x dx
Second pass: u = cos x, dv = eˣ dx → eˣ cos x + ∫ eˣ sin x dx
Substituting back: ∫ eˣ sin x dx = eˣ sin x − eˣ cos x − ∫ eˣ sin x dx
Move the integral to the left: 2 ∫ eˣ sin x dx = eˣ(sin x − cos x)
Result: (1/2) eˣ (sin x − cos x) + C
The reason you move the integral to the other side: after two rounds of IBP, the original integral shows up again with the opposite sign. Rather than looping forever, you treat it as an algebraic unknown and solve for it.
General formula (function notation): ∫ f(x)g'(x) dx = f(x)g(x) − ∫ g(x)f'(x) dx
Shorthand notation: ∫ u dv = uv − ∫ v du
LIATE priority for choosing u: Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential
Cycling result (memorise this pattern): ∫ eˣ sin x dx = (1/2) eˣ (sin x − cos x) + C ∫ eˣ cos x dx = (1/2) eˣ (sin x + cos x) + C
Integration by parts is how engineers and physicists compute quantities like the centre of mass of a non-uniform rod (∫ x · density(x) dx), the expected value of a continuous random variable in probability and statistics, and the work done by a variable force. In signal processing, Fourier transforms rely on IBP to move between the time and frequency domains. Any time you see a product of two different types of function under an integral sign in an applied setting, IBP is likely the method behind the calculation.
Students often set dv = ln x. Do not do this. Logarithms are almost always u, because differentiating ln x gives 1/x (simpler), while integrating ln x requires IBP itself.
Students forget that dv must include the dx. If you write dv = eˣ instead of dv = eˣ dx, the notation breaks and you will lose track of what to integrate.
When using IBP on definite integrals, students sometimes apply the bounds to the uv term but forget to apply them to the remaining ∫ v du term as well. Both pieces need the bounds.
In cycling problems, students apply IBP a third and fourth time instead of recognising the original integral on the right-hand side. If the same integral reappears, stop and solve algebraically.
⚠️ IBP is tested heavily. Expect at least one IBP problem on every Calc 2 exam, often combined with definite integral evaluation.
⚠️ The cycling technique (eˣ sin x or eˣ cos x) is a classic exam question. Know the pattern cold.
⚠️ Repeated IBP with polynomials times exponentials is a favourite for showing whether students can carry through multiple rounds without sign errors.
⚠️ The "dv = dx" trick for ∫ ln x dx catches students off guard. Remember that you can always set dv = 1 dx when there is no obvious second factor.
True or False: In ∫ x eˣ dx, you should let u = eˣ and dv = x dx.
False. Let u = x (algebraic, simplifies on differentiation) and dv = eˣ dx.
Fill in the blank: ∫ u dv = uv − ______.
∫ v du
True or False: When IBP produces the original integral on the right-hand side, you should apply IBP a third time.
False. Move the integral to the left-hand side and solve algebraically.
Fill in the blank: ∫ ln x dx = x ln x − ______ + C.
x
True or False: For definite integrals using IBP, you only apply the bounds to the uv term.
False. You must evaluate both the uv term and the ∫ v du term at the bounds.
Q: Evaluate ∫ x cos x dx.
A: Let u = x, dv = cos x dx. Then du = dx, v = sin x. IBP gives x sin x − ∫ sin x dx = x sin x + cos x + C.
Q: Evaluate ∫ x² eˣ dx.
A: Apply IBP twice. First pass: u = x², dv = eˣ dx → x² eˣ − 2 ∫ x eˣ dx. Second pass on ∫ x eˣ dx: u = x, dv = eˣ dx → x eˣ − eˣ. Combine: x² eˣ − 2(x eˣ − eˣ) = eˣ(x² − 2x + 2) + C.
Q: Evaluate ∫ eˣ cos x dx.
A: Apply IBP twice, then solve for the integral algebraically. Result: (1/2) eˣ (sin x + cos x) + C.
Q: Evaluate ∫ from 0 to 1 of x e⁻ˣ dx.
A: u = x, dv = e⁻ˣ dx, v = −e⁻ˣ, du = dx. Antiderivative: −x e⁻ˣ − e⁻ˣ = −e⁻ˣ(x + 1). Evaluate: −e⁻¹(2) − (−1) = 1 − 2/e.
Q: Why does the cycling method work for ∫ eˣ sin x dx?
A: Both sin x and eˣ reproduce themselves under differentiation and integration. After two rounds of IBP, the original integral appears on the right with the opposite sign. Moving it to the left gives 2 ∫ eˣ sin x dx = eˣ(sin x − cos x), so you divide by 2.
IBP connects directly to u-substitution: both are reverse-engineering the chain rule and product rule, respectively. Many Calc 2 problems require you to decide which one to use first, or to combine them (IBP followed by a substitution to finish the remaining integral).
This technique is essential preparation for reduction formulas (∫ sinⁿ x dx, ∫ cosⁿ x dx), which are derived by applying IBP and then using trig identities.
In differential equations, IBP appears when solving for particular solutions using variation of parameters and when computing Laplace transforms, both of which you will meet if you continue to Calc 3 or an ODE course.
integration by parts, IBP, LIATE, ILATE, tabular method, tabular integration, product rule for integrals, cycling integral, ∫ u dv, uv minus integral v du, Calc 2 integration techniques, integral of x e^x, integral of ln x, integral of e^x sin x, repeated integration by parts, definite integral IBP, Fundamental Theorem of Calculus with IBP, reduction formula, UT Austin Calc 2, integral calculus