Integration by Parts

Quick answer

Integration by parts reverses the product rule. From (uv)′ = u′v + uv′ comes ∫u dv = uv − ∫v du, which trades one integral for another. Choose u to be a factor that gets simpler when differentiated, such as a power of x or ln x, and dv a factor you can integrate, such as eˣ, sin x or cos x. Some integrals need parts twice; some come back to where they started, and then the equation can be solved for the integral.

What you'll learn

Derive integration by parts from the product rule
Choose u and dv so the new integral is simpler
Apply parts repeatedly, and solve for an integral that returns
Evaluate definite integrals by parts

The product rule, backward

Substitution undoes the chain rule. Integration by parts undoes the product rule. Start from $(uv)' = u'v + uv'$ and integrate both sides:

uv = \int v\,u'\,dx + \int u\,v'\,dx

With $du = u'\,dx$ and $dv = v'\,dx$ , solve for the second integral:

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

To use it, split the integrand into two factors: $u$ , which you will differentiate, and $dv$ , which you will integrate.

Why parts helps

The formula trades $\int u\,dv$ for $\int v\,du$ . The trade moves the derivative from one factor to the other: $u$ gets differentiated and $dv$ gets integrated. If differentiating $u$ makes it simpler, as it turns $x$ into $1$ , the new integral is easier than the old. Choose $u$ so that differentiating it simplifies the problem.

The formula has a picture. As $x$ runs along, the point $(v, u)$ traces a curve. When $u$ and $v$ both start at $0$ , the area under the curve is $\int u\,dv$ and the area to its left is $\int v\,du$ . Together they fill a rectangle of area $uv$ .

the curve traced by (v, u)

Two areas that fill a rectangle

The LIATE order is a reliable guide to choosing $u$ : take the first factor on the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Factors late in the list have simple antiderivatives, so they make good choices for $dv$ .

Worked examples

Common mistakes

Practice problems

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

Answer

$x\sin x + \cos x + C$

Full solution

$u = x$ , $dv = \cos x\,dx$ , $v = \sin x$ : $x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$ .
Find $\int x e^{2x}\,dx$ .

Answer

$\tfrac{x}{2}e^{2x} - \tfrac{1}{4}e^{2x} + C$

Full solution

$u = x$ , $v = \tfrac{1}{2}e^{2x}$ : $\tfrac{x}{2}e^{2x} - \int \tfrac{1}{2}e^{2x}\,dx = \tfrac{x}{2}e^{2x} - \tfrac{1}{4}e^{2x} + C$ .
Find $\int x\ln x\,dx$ .

Answer

$\tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C$

Full solution

LIATE puts the logarithm first: $u = \ln x$ , $dv = x\,dx$ , $v = \tfrac{x^2}{2}$ . Then $\tfrac{x^2}{2}\ln x - \int \tfrac{x^2}{2} \cdot \tfrac{1}{x}\,dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C$ .
Find $\int \arctan x\,dx$ .

Answer

$x\arctan x - \tfrac{1}{2}\ln(1 + x^2) + C$

Full solution

$u = \arctan x$ , $dv = dx$ : $x\arctan x - \int \tfrac{x}{1 + x^2}\,dx$ . Substitute $w = 1 + x^2$ in the last integral to get $\tfrac{1}{2}\ln(1 + x^2)$ .
Find $\int x^2 e^x\,dx$ .

Answer

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

Full solution

$u = x^2$ : $x^2e^x - \int 2xe^x\,dx$ . By Example 1, $\int 2xe^x\,dx = 2xe^x - 2e^x$ . So the answer is $x^2e^x - 2xe^x + 2e^x + C$ .
Evaluate $\int_0^1 x e^x\,dx$ .

Answer

$1$

Full solution

$\big[x e^x - e^x\big]_0^1 = (e - e) - (0 - 1) = 1$ .
Evaluate $\int_1^e \ln x\,dx$ .

Answer

$1$

Full solution

$\big[x\ln x - x\big]_1^e = (e - e) - (0 - 1) = 1$ .
Find $\int e^x\cos x\,dx$ .

Answer

$\tfrac{e^x}{2}(\sin x + \cos x) + C$

Full solution

Call it $J$ . Parts with $u = \cos x$ : $J = e^x\cos x + \int e^x\sin x\,dx$ . Parts again with $u = \sin x$ : $\int e^x\sin x\,dx = e^x\sin x - J$ . So $J = e^x\cos x + e^x\sin x - J$ , and $2J = e^x(\sin x + \cos x)$ .
Find $\int x\sec^2 x\,dx$ .

Answer

$x\tan x + \ln\lvert \cos x \rvert + C$

Full solution

$u = x$ , $dv = \sec^2 x\,dx$ , $v = \tan x$ : $x\tan x - \int \tan x\,dx = x\tan x + \ln\lvert \cos x \rvert + C$ , using $\int \tan x\,dx = -\ln\lvert \cos x \rvert + C$ .
A student tries $\int x e^x\,dx$ with $u = e^x$ and $dv = x\,dx$ , gets $\tfrac{x^2}{2}e^x - \int \tfrac{x^2}{2}e^x\,dx$ , and concludes the integral cannot be done. What went wrong?

Hint

Which factor gets simpler when differentiated?

Answer

The choice of $u$ . With $u = x$ and $dv = e^x\,dx$ , the answer is $x e^x - e^x + C$ .

Full solution

Differentiating $e^x$ leaves it unchanged, while integrating $x$ raises its power, so the student’s new integral is harder than the original.

Differentiating $x$ gives $1$ , so taking $u = x$ leaves $\int e^x\,dx$ , which is immediate.

Frequently asked questions

What is the integration by parts formula?

∫u dv = uv − ∫v du. For definite integrals, the uv term is evaluated between the limits.

How do I choose u?

Pick the factor that gets simpler when you differentiate it. The LIATE order — logarithmic, inverse trig, algebraic, trigonometric, exponential — lists the usual first choices for u.

What if the new integral is harder than the original?

Swap the choices. In ∫x eˣ dx, taking u = eˣ raises the power of x; taking u = x removes it.

What if integrating by parts brings back the original integral?

Treat the integral as an unknown I. The equation I = (something) − I can be solved: I = (something)/2, plus C.

How do I integrate ln x?

Use parts with u = ln x and dv = dx: ∫ln x dx = x ln x − x + C.

What to learn next

Integration by Partial FractionsIntegrate rational functions by splitting them into simpler fractions: distinct linear factors, finding the constants, repeated factors, and dividing first.

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