Integration by Parts: Formula & Examples

Integration by parts is a method for integrating a product of two functions. It uses the formula $\int u\,dv = uv - \int v\,du$ . Choose u to be the part that gets simpler when you differentiate it, and dv to be the rest. The new integral, $\int v\,du$ , should be easier than the one you started with.

Before this lesson: U-Substitution, Product Rule and Antiderivatives and Indefinite Integrals

Where the formula comes from

Integration by parts is the product rule run backward. For two functions u and v, the product rule says (uv)′ = u′v + uv′. Integrate both sides: uv = ∫v du + ∫u dv. Then solve for ∫u dv.

The point is to trade one integral for another. If you pick u well, ∫v du has a simpler integrand than ∫u dv, and you can finish it with a basic rule.

A two-by-two box. Top row: u = x and dv = cos x dx. Bottom row: du = dx and v = sin x. Side labels say to differentiate u and integrate dv. Below: the integral of u dv equals uv minus the integral of v du, so the integral of x cos x dx = x sin x minus the integral of sin x dx. A strip lists LIATE: log, inverse trig, algebraic, trig, exponential. — Set up integration by parts with a box: differentiate u, integrate dv, then use uv − ∫v du. LIATE helps you choose u.

Integration by parts formula

\int u\,dv = uv - \int v\,du

For a definite integral:

\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du

Choosing u with LIATE: pick u from the type of function that comes first in this list. Then dv is everything else, including dx.

Logarithmic: ln x
Inverse trig: arctan x, arcsin x
Algebraic: x, x², 3x + 1
Trig: sin x, cos x
Exponential: eˣ, e²ˣ

Steps:

Choose u and dv.
Find du by differentiating u, and v by integrating dv. You can leave + C off v.
Write uv − ∫v du.
Find the new integral. Use parts again if you need to.
Add + C at the end, or evaluate at the limits for a definite integral.

Integration by parts examples

Example 1: x times a trig function

Find $\int x\cos x\,dx$ .

u = x, \; dv = \cos x\,dx

LIATE: x is algebraic and comes before trig, so u = x.

du = dx, \; v = \sin x

Differentiate u and integrate dv.

\int x\cos x\,dx = x\sin x - \int \sin x\,dx

Use uv − ∫v du.

= x\sin x - (-\cos x) + C

An antiderivative of sin x is −cos x (see antiderivatives).

= x\sin x + \cos x + C

Subtracting a negative adds.

Answer: $x\sin x + \cos x + C$ . Check: by the product rule, the derivative is sin x + x cos x − sin x = x cos x.

Example 2: x times an exponential

Find $\int x e^{2x}\,dx$ .

u = x, \; dv = e^{2x}\,dx

Algebraic comes before exponential.

du = dx, \; v = \dfrac{1}{2}e^{2x}

Integrate e²ˣ with u-substitution (see u-substitution).

\int x e^{2x}\,dx = \dfrac{x}{2}e^{2x} - \int \dfrac{1}{2}e^{2x}\,dx

Use uv − ∫v du.

= \dfrac{x}{2}e^{2x} - \dfrac{1}{4}e^{2x} + C

Integrate (1/2)e²ˣ.

= \dfrac{e^{2x}}{4}(2x - 1) + C

Factor out e²ˣ/4.

Answer: $\dfrac{e^{2x}}{4}(2x - 1) + C$

Example 3: The integral of ln x

Find $\int \ln x\,dx$ .

u = \ln x, \; dv = dx

Logarithms come first in LIATE. dv is just dx.

du = \dfrac{1}{x}\,dx, \; v = x

Differentiate ln x (see derivatives of logarithmic functions); integrate dx.

\int \ln x\,dx = x\ln x - \int x\cdot\dfrac{1}{x}\,dx

Use uv − ∫v du.

= x\ln x - \int 1\,dx

Simplify: x times 1/x is 1.

= x\ln x - x + C

Integrate 1.

Answer: $x\ln x - x + C$ . Even a single function can use parts, with dv = dx.

Example 4: Using parts twice

Find $\int x^2 e^x\,dx$ .

u = x^2, \; dv = e^x\,dx

Algebraic before exponential.

du = 2x\,dx, \; v = e^x

Differentiate and integrate.

\int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx

First round of parts. The power of x dropped by one.

\int 2x e^x\,dx = 2x e^x - \int 2e^x\,dx

Second round: u = 2x, dv = eˣ dx, so du = 2 dx and v = eˣ.

= 2x e^x - 2e^x

Integrate 2eˣ.

\int x^2 e^x\,dx = x^2 e^x - (2x e^x - 2e^x) + C

Put the second result into the first.

= e^x(x^2 - 2x + 2) + C

Distribute the minus sign and factor out eˣ.

Answer: $e^x(x^2 - 2x + 2) + C$ . Each round lowers the power of x, so a power of n needs n rounds.

Example 5: Distance traveled

A particle moves with velocity $v(t) = t e^{-t}$ meters per second. How far does it travel from t = 0 to t = 2?

u = t, \; dv = e^{-t}\,dt

Algebraic before exponential.

du = dt, \; v = -e^{-t}

An antiderivative of e^(−t) is −e^(−t).

\int_0^2 t e^{-t}\,dt = \Big[-t e^{-t}\Big]_0^2 - \int_0^2 -e^{-t}\,dt

Definite form of parts.

= -2e^{-2} + \int_0^2 e^{-t}\,dt

Evaluate uv; the minus signs cancel in the integral.

= -2e^{-2} + \Big[-e^{-t}\Big]_0^2

Integrate e^(−t).

= -2e^{-2} - e^{-2} + 1

Plug in the limits: −e^(−2) − (−1).

= 1 - 3e^{-2} \approx 0.594

Combine like terms; e^(−2) ≈ 0.1353.

Answer: about 0.594 meters. The velocity is never negative here, so the distance equals the displacement (see displacement and total distance).

Common mistakes

Mistake	Example of the error	Fix
Choosing u backward	For $\int x e^x\,dx$ , u = eˣ gives $\dfrac{x^2}{2}e^x - \int \dfrac{x^2}{2}e^x\,dx$ , which is harder	Follow LIATE: u = x, so du = dx and the new integral is simpler.
Sign error in uv − ∫v du	$\int x\cos x\,dx = x\sin x + \int \sin x\,dx$	The formula has a minus: $x\sin x - \int \sin x\,dx$ .
Differentiating dv instead of integrating it	dv = cos x dx, then v = −sin x	v is an antiderivative of cos x: v = sin x.
Dropping the minus sign in the second round	$x^2e^x - 2xe^x - 2e^x$	Keep the parentheses: $x^2e^x - (2xe^x - 2e^x)$ , which gives + 2eˣ.
Forgetting uv at the limits	$\int_0^2 t e^{-t}\,dt = \int_0^2 e^{-t}\,dt$	Evaluate uv from a to b too: $\Big[-te^{-t}\Big]_0^2 = -2e^{-2}$ .

Related methods

Tabular integration

When u is a polynomial and dv can be integrated again and again, make a table. Differentiate u down to 0 in one column, integrate dv in the other, and multiply along diagonals with alternating signs (+, −, +, …). It gives Example 4’s answer in one step.

Integrals that come back

For $\int e^x\sin x\,dx$ , two rounds of parts bring back the original integral. Call it I, write the equation, and solve for I like an algebra problem. Practice problem 9 shows how.

Other techniques

If the integrand is an inside function times its derivative, u-substitution is faster. Powers of sine and cosine use the next lesson, Trigonometric Integrals.

Practice problems

Find $\int x\sin x\,dx$ .
Find $\int x e^x\,dx$ .
Find $\int x\ln x\,dx$ .
Find $\int (2x + 1)e^x\,dx$ .
Find $\int \arctan x\,dx$ .
Find $\int x^2\sin x\,dx$ .
Evaluate $\int_0^1 x e^{-x}\,dx$ .
Evaluate $\int_1^e \ln x\,dx$ .
Find $\int e^x\sin x\,dx$ .
Water flows into a tank at $r(t) = 20t e^{-t/2}$ liters per minute. How much water flows in during the first 4 minutes?

Show answers

u = x, \; dv = \sin x\,dx, \; du = dx, \; v = -\cos x

Choose u and dv; find du and v.

\int x\sin x\,dx = -x\cos x - \int -\cos x\,dx

Use uv − ∫v du.

= -x\cos x + \sin x + C

Integrate cos x.

Answer: $-x\cos x + \sin x + C$

u = x, \; dv = e^x\,dx, \; du = dx, \; v = e^x

Choose u and dv.

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

Use uv − ∫v du and integrate.

Answer: $x e^x - e^x + C$

u = \ln x, \; dv = x\,dx, \; du = \dfrac{1}{x}\,dx, \; v = \dfrac{x^2}{2}

Logs come first in LIATE.

\int x\ln x\,dx = \dfrac{x^2}{2}\ln x - \int \dfrac{x^2}{2}\cdot\dfrac{1}{x}\,dx

Use uv − ∫v du.

= \dfrac{x^2}{2}\ln x - \int \dfrac{x}{2}\,dx = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C

Simplify and integrate.

Answer: $\dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$

u = 2x + 1, \; dv = e^x\,dx, \; du = 2\,dx, \; v = e^x

Choose u and dv.

\int (2x + 1)e^x\,dx = (2x + 1)e^x - \int 2e^x\,dx

Use uv − ∫v du.

= (2x + 1)e^x - 2e^x + C = (2x - 1)e^x + C

Integrate and factor out eˣ.

Answer: $(2x - 1)e^x + C$

u = \arctan x, \; dv = dx, \; du = \dfrac{1}{1 + x^2}\,dx, \; v = x

Inverse trig comes early in LIATE.

\int \arctan x\,dx = x\arctan x - \int \dfrac{x}{1 + x^2}\,dx

Use uv − ∫v du.

= x\arctan x - \dfrac{1}{2}\ln(1 + x^2) + C

The top is half the derivative of the bottom (see integrals that give natural logarithms).

Answer: $x\arctan x - \dfrac{1}{2}\ln(1 + x^2) + C$

\int x^2\sin x\,dx = -x^2\cos x - \int -2x\cos x\,dx

Round 1: u = x², dv = sin x dx, so v = −cos x.

= -x^2\cos x + 2\int x\cos x\,dx

Simplify the signs.

= -x^2\cos x + 2(x\sin x + \cos x) + C

Round 2 is Example 1.

= -x^2\cos x + 2x\sin x + 2\cos x + C

Distribute the 2.

Answer: $-x^2\cos x + 2x\sin x + 2\cos x + C$

\int_0^1 x e^{-x}\,dx = \Big[-x e^{-x}\Big]_0^1 + \int_0^1 e^{-x}\,dx

u = x, dv = e^(−x) dx, so v = −e^(−x).

= -e^{-1} + \Big[-e^{-x}\Big]_0^1 = -e^{-1} - e^{-1} + 1

Evaluate both parts.

= 1 - \dfrac{2}{e} \approx 0.264

Combine like terms.

Answer: $1 - \dfrac{2}{e} \approx 0.264$

\int_1^e \ln x\,dx = \Big[x\ln x - x\Big]_1^e

Use the result of Example 3.

= (e\ln e - e) - (1\ln 1 - 1) = (e - e) - (0 - 1) = 1

ln e = 1 and ln 1 = 0.

Answer: 1

I = \int e^x\sin x\,dx = -e^x\cos x + \int e^x\cos x\,dx

Round 1: u = eˣ, dv = sin x dx, v = −cos x.

\int e^x\cos x\,dx = e^x\sin x - \int e^x\sin x\,dx

Round 2: u = eˣ, dv = cos x dx, v = sin x. The original integral is back.

2I = e^x\sin x - e^x\cos x

Substitute: I = −eˣcos x + eˣsin x − I, then add I to both sides.

I = \dfrac{e^x}{2}(\sin x - \cos x) + C

Divide by 2 and add + C.

Answer: $\dfrac{e^x}{2}(\sin x - \cos x) + C$

10.

u = t, \; dv = e^{-t/2}\,dt, \; du = dt, \; v = -2e^{-t/2}

Choose u and dv; integrate e^(−t/2).

\int_0^4 t e^{-t/2}\,dt = \Big[-2t e^{-t/2}\Big]_0^4 + 2\int_0^4 e^{-t/2}\,dt

Definite form of parts.

= -8e^{-2} + 2\Big[-2e^{-t/2}\Big]_0^4 = -8e^{-2} - 4e^{-2} + 4

Evaluate both parts.

= 4 - 12e^{-2}

Combine like terms.

20(4 - 12e^{-2}) = 80 - 240e^{-2} \approx 47.5

Multiply by the 20 in r(t).

Answer: about 47.5 liters

Integration by parts FAQ

What is the integration by parts formula?

$\int u\,dv = uv - \int v\,du$ . It comes from the product rule for derivatives.

How do you choose u in integration by parts?

Use LIATE: logarithmic, inverse trig, algebraic, trig, exponential. Pick u from the type that comes first. The goal is a u that gets simpler when you differentiate it.

When should you use integration by parts?

Use it for products of different kinds of functions, like x eˣ, x sin x or x² ln x, and for ln x or arctan x alone. If the integrand is a function of an inside piece times that piece’s derivative, try u-substitution first.

What if the integral keeps coming back?

That happens with products like eˣ sin x. After two rounds, the original integral appears again. Treat it as an unknown and solve the equation for it.

Summary

$\int u\,dv = uv - \int v\,du$ , the product rule run backward.
Choose u with LIATE; dv is everything else.
Differentiate u to get du, and integrate dv to get v.
Repeat parts when a power of x is still there.
For definite integrals, evaluate uv at both limits.
If the original integral comes back, solve for it.

Previous course: Calculus 1 Index

Next lesson: Trigonometric Integrals

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