Integration by Parts

Reverse a product derivative

Integrating $(uv)'=u'v+uv'$ and rearranging yields

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

The method exchanges one integral for another. A useful choice makes $u$ simpler after differentiation and makes $dv$ easy to integrate. There is no universal ordering trick that guarantees the best choice.

For definite integrals the same identity reads $\int_a^buv'=[uv]_a^b-\int_a^bu'v$ . Keep the boundary term and the same bounds on the new integral. If the integrals are improper, apply the identity first on a finite safe interval and then take the required limit.

Visual guide

VISUAL GUIDETwo complementary areas fill a rectangle

y = eˣ

For the inverse pair y = eˣ and x = ln y between x = 0 and 1, ∫₀¹eˣ dx + ∫₁ᵉ ln y dy = e. The shaded region under eˣ and the remaining region to its left fill the rectangle of width 1 and height e. This is a geometric instance of integration by parts.

Worked example: polynomial times exponential

For $\int xe^x dx$ , choose $u=x$ and $dv=e^x dx$ . Then $du=dx$ , $v=e^x$ , and

\int xe^x dx=xe^x-\int e^x dx=xe^x-e^x+C.

The polynomial lost a degree and the remaining integral is elementary. Differentiating the result gives $e^x+xe^x-e^x=xe^x$ , which checks both the product term and the subtraction.

For $x^2e^x$ , repeating this choice gives $e^x(x^2-2x+2)+C$ . Repetition works because successive polynomial derivatives eventually vanish. Choosing the polynomial as $dv$ would instead increase its degree and usually make the remaining problem worse.

Worked example: a hidden product

Write $\int\ln x\,dx$ as $\int(\ln x)\cdot1\,dx$ on $x>0$ . Choose $u=\ln x$ , $dv=dx$ . This gives $x\ln x-\int1dx=x\ln x-x+C$ . The logarithm is hard to integrate directly but easy to differentiate.

Now consider $I=\int e^x\cos x\,dx$ . Applying parts twice produces $I=e^x\cos x+e^x\sin x-I$ . Solve the algebraic equation to obtain $I=\tfrac12e^x(\sin x+\cos x)+C$ . A recurring original integral can be useful; it is not necessarily a sign of failure.

Choosing and checking a method

Look for a product where one factor becomes simpler after differentiation. Substitution is often better for a composition accompanied by its derivative, such as $x e^{x^2}$ . Neither method should be applied mechanically based on a single visible symbol.

Use one final arbitrary constant for an indefinite result; intermediate choices of constants for $v$ cancel or are absorbed. For definite results, no arbitrary constant remains.

Practice

Evaluate $\int_0^1 xe^x dx$ .
Integrate $x\cos x$ .
Integrate $\arctan x$ .

Show worked solutions

$[xe^x-e^x]_0^1=0-(-1)=1$ .
Choose $u=x$ , $dv=\cos x\,dx$ . The result is $x\sin x+\cos x+C$ .
With $u=\arctan x$ , $dv=dx$ , obtain $x\arctan x-\int x/(1+x^2)dx=x\arctan x-\tfrac12\ln(1+x^2)+C$ .