What Is Integration by Parts?
Integration by parts is a technique for integrating products of functions that cannot be handled by basic rules or u-substitution. It is the integration counterpart of the product rule for derivatives.
The product rule says d/dx [u · v] = u · v' + v · u'. Rearranging and integrating both sides gives the integration by parts formula:
∫ u dv = uv − ∫ v duThe idea is simple: if the original integral ∫ u dv is hard, maybe the new integral ∫ v du is easier. You are trading one integral for another, hopefully simpler, one.
The formula requires you to split the integrand into two pieces: one piece you call u (which you will differentiate to get du) and the rest you call dv (which you will integrate to get v). Choosing wisely is the key skill — a poor choice can make the integral harder instead of easier.
Integration by parts is typically introduced in Calculus II (or AP Calculus BC) and is one of the most frequently tested techniques on exams. It handles integrals like ∫ x·eˣ dx, ∫ x·sin(x) dx, ∫ ln(x) dx, and ∫ eˣ·cos(x) dx that none of the simpler methods can touch.
How to Choose u and dv — The LIATE Rule
The success of integration by parts hinges on choosing the right u and dv. A bad choice leads to a more complicated integral; a good choice simplifies it. The LIATE rule is a widely used guideline that tells you what to pick for u by priority.
ln(x), log(x)). I — Inverse trig functions (arcsin(x), arctan(x)). A — Algebraic functions (x, x², x³, polynomials). T — Trigonometric functions (sin(x), cos(x)). E — Exponential functions (eˣ, 2ˣ).The function that appears earliest in LIATE should be your choice for u, and everything else goes into dv.
Examples of LIATE in action
For example, in ∫ x·eˣ dx, x is algebraic (A) and eˣ is exponential (E). Since A comes before E, let u = x and dv = eˣ dx.
In ∫ ln(x)·x² dx, ln(x) is logarithmic (L) and x² is algebraic (A). Since L comes before A, let u = ln(x) and dv = x² dx.
The LIATE rule is a guideline, not a law — it works in the vast majority of cases, but occasionally you may need to try a different split. The underlying principle is: choose u to be the function that becomes simpler when differentiated, and dv to be something you can easily integrate.
Step-by-Step Example: ∫ x·eˣ dx
Let's work through the classic example ∫ x·eˣ dx step by step.
- Choose u and dv
Using LIATE,
xis algebraic andeˣis exponential, so letu = xanddv = eˣ dx. - Find du and v
Differentiate
u:du = dx. Integratedv:v = eˣ. - Apply the formula
∫ x·eˣ dx = uv − ∫ v du = x·eˣ − ∫ eˣ dx. - Evaluate the remaining integral
∫ eˣ dx = eˣ. - Combine
∫ x·eˣ dx = x·eˣ − eˣ + C = eˣ(x − 1) + C.
∫ x·eˣ dx = eˣ(x − 1) + CYou can verify this by differentiating: d/dx [eˣ(x − 1)] = eˣ(x − 1) + eˣ(1) = eˣ·x − eˣ + eˣ = x·eˣ.
The strategy worked because differentiating u = x reduced it to a constant (du = dx), and integrating dv = eˣ dx was trivial (v = eˣ). The new integral ∫ eˣ dx was much simpler than the original.
When You Need to Apply It Twice
Some integrals require two rounds of integration by parts. The classic example is ∫ x²·eˣ dx.
- Round 1
Let
u = x²,dv = eˣ dx. Thendu = 2x dx,v = eˣ. The formula gives∫ x²·eˣ dx = x²·eˣ − ∫ 2x·eˣ dx. The new integral∫ 2x·eˣ dxis simpler (the power ofxdropped from 2 to 1), but it still requires integration by parts. - Round 2
For
∫ 2x·eˣ dx, letu = 2x,dv = eˣ dx. Thendu = 2 dx,v = eˣ. So∫ 2x·eˣ dx = 2x·eˣ − ∫ 2eˣ dx = 2x·eˣ − 2eˣ. - Combine everything
∫ x²·eˣ dx = x²·eˣ − (2x·eˣ − 2eˣ) + C = eˣ(x² − 2x + 2) + C.
eˣ(x² − 2x + 2) + CIn general, if u is a polynomial of degree n and dv involves eˣ or a trig function, you will need n rounds of integration by parts. Each round reduces the polynomial degree by one.
The boomerang technique
Another case requiring two applications is ∫ eˣ·sin(x) dx. Here, integration by parts twice brings you back to the original integral, and you solve for it algebraically:
I = eˣ·sin(x) − eˣ·cos(x) − I
so 2I = eˣ(sin(x) − cos(x))
I = ½·eˣ(sin(x) − cos(x)) + CThis "boomerang" technique is a favorite on exams.
Integration by Parts for ln(x) and Inverse Trig
Integration by parts is the standard method for integrals involving ln(x) or inverse trig functions, because these functions are easy to differentiate but hard (or impossible) to integrate directly.
This looks like a single function, but you can write it as ∫ ln(x) · 1 dx.
Let u = ln(x) and dv = dx. Then du = (1/x) dx and v = x.
The formula gives ∫ ln(x) dx = x·ln(x) − ∫ x · (1/x) dx = x·ln(x) − ∫ 1 dx = x·ln(x) − x + C.
∫ ln(x) dx = x·ln(x) − x + CThis is a result worth memorizing.
Write as ∫ arctan(x) · 1 dx.
Let u = arctan(x), dv = dx. Then du = 1/(1 + x²) dx and v = x.
The formula gives ∫ arctan(x) dx = x·arctan(x) − ∫ x/(1 + x²) dx.
The remaining integral is a standard substitution: let w = 1 + x², dw = 2x dx, so ∫ x/(1 + x²) dx = ½·ln|1 + x²|.
∫ arctan(x) dx = x·arctan(x) − ½·ln(1 + x²) + CThe pattern for all these integrals is the same: let u be the function you know how to differentiate, and let dv = dx.
Definite Integrals and the Boundary Term
For definite integrals, the integration by parts formula becomes:
∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v duThe boundary term [uv]ₐᵇ means you evaluate u·v at the upper limit and subtract its value at the lower limit.
Evaluate ∫₀¹ x·eˣ dx.
Using the result from earlier, ∫ x·eˣ dx = eˣ(x − 1).
Evaluate at the bounds: [eˣ(x − 1)]₀¹ = e¹(1 − 1) − e⁰(0 − 1) = 0 − (−1) = 1.
∫₀¹ x·eˣ dx = 1Evaluate ∫₁ᵉ ln(x) dx.
Using ∫ ln(x) dx = x·ln(x) − x:
[x·ln(x) − x]₁ᵉ = (e·ln(e) − e) − (1·ln(1) − 1) = (e − e) − (0 − 1) = 0 + 1 = 1.
∫₁ᵉ ln(x) dx = 1When applying integration by parts to definite integrals, you can either find the antiderivative first and then evaluate at the bounds, or carry the bounds through every step. Both approaches give the same answer. Most students find it cleaner to find the full antiderivative first, then plug in the limits at the end.
Practice Problems
Evaluate ∫ x·cos(x) dx.
Answer: Let u = x, dv = cos(x) dx. Then du = dx, v = sin(x).
∫ x·cos(x) dx = x·sin(x) − ∫ sin(x) dx = x·sin(x) + cos(x) + C
x·sin(x) + cos(x) + CEvaluate ∫ x²·sin(x) dx.
Hint: Apply integration by parts twice.
−x²·cos(x) + 2x·sin(x) + 2cos(x) + CEvaluate ∫ ln(x)/x² dx.
Hint: Let u = ln(x), dv = x⁻² dx.
−ln(x)/x − 1/x + C = −(ln(x) + 1)/x + CEvaluate ∫₀^π x·sin(x) dx.
Answer: Let u = x, dv = sin(x) dx.
∫ x·sin(x) dx = −x·cos(x) + sin(x)
Evaluate: [−x·cos(x) + sin(x)]₀^π = (−π·(−1) + 0) − (0 + 0) = π.
πEvaluate ∫ eˣ·cos(x) dx.
Hint: Apply integration by parts twice and solve for the integral algebraically.
½·eˣ(sin(x) + cos(x)) + CIf you want to check your integration by parts work or see the full breakdown of which function to choose for u and dv, scan the problem with Solver AI for a complete step-by-step solution.