Integration by Parts¶

Proposition¶

Let $f$ and $g$ be differentiable functions on $[a, b]$ with $f'$ and $g'$ Riemann integrable. Then:

\int_a^b f(x)g'(x)\,\mathrm{d}x =f(b)g(b)-f(a)g(a) -\int_a^b f'(x)g(x)\,\mathrm{d}x,

or

\int_a^b f(x)\,\mathrm{d}g(x) =f(b)g(b)-f(a)g(a) -\int_a^b g(x)\,\mathrm{d}f(x).

If we consider the indefinite integral, we get:

\int f\,\mathrm{d}g = fg -\int g\,\mathrm{d}f+C.

Interpretation¶

Integration by parts is the product rule read backwards. Since $(fg)'=f'g+fg'$ $f g^{'}$ , integrating gives $fg$ back, so an integral of $fg'$ can be traded for an integral of $f'g$ plus a boundary term. This helps when differentiating $f$ makes it simpler while $g'$ is easy to integrate.

An increasing curve in the (u, v) plane dividing a rectangle into two regions, the integral of u dv on one side and the integral of v du on the other, which together give u2 v2 minus u1 v1.

Put $u=f(x)$ and $v=g(x)$ $g (x)$ . As $x$ runs from $a$ to $b$ , the point $(u,v)$ traces the curve from $(u_1,v_1)$ to $(u_2,v_2)$ . The area below the curve is $\int v\,\mathrm{d}u$ and the area to its left is $\int u\,\mathrm{d}v$ ; together they fill the large rectangle minus the small one, so $\int u\,\mathrm{d}v+\int v\,\mathrm{d}u=u_2v_2-u_1v_1$ $u_{1} v_{1}$ . With $u_1=f(a)$ $f (a)$ , $u_2=f(b)$ and so on, this is the second form of the formula. The picture assumes increasing positive $f$ and $g$ ; the formula does not.

The differential notation $\mathrm{d}g=g'(x)\,\mathrm{d}x$ rewrites the same product-rule integral. It identifies which factor is differentiated; it does not make $\mathrm{d}g$ an unrelated integration variable.

The choice of $f$ and $g'$ decides whether the method helps. For $\int x\mathrm{e}^x\,\mathrm{d}x$ , taking $f(x)=x$ turns the new integrand into $\mathrm{e}^x$ . The opposite choice $f(x)=\mathrm{e}^x$ $e^{x}$ , $g'(x)=x$ is a common slip: it leads to $\frac{x^2}{2}\mathrm{e}^x-\int\frac{x^2}{2}\mathrm{e}^x\,\mathrm{d}x$ $\int \frac{x ^{2}}{2} e^{x} d x$ , which is harder than the original.

The same step proves the recursion $\Gamma(x+1)=x\,\Gamma(x)$ of the gamma function. If the integrand is not defined at an endpoint, first work away from that endpoint and then take a limit, as in the second Example; this procedure also applies to improper integrals.

Key Results¶

Integration by parts follows directly from the product rule.
The method can be applied to both definite and indefinite integrals.
Appropriate choice of functions is essential for simplification.

Examples¶

Choose $f(x)=x$ and $g'(x)=\mathrm{e}^x$ $e^{x}$ . Then $f'(x)=1$ and $g(x)=\mathrm{e}^x$ $e^{x}$ , so $\int x\mathrm{e}^x\,\mathrm{d}x=x\mathrm{e}^x-\mathrm{e}^x+C=(x-1)\mathrm{e}^x+C$ $C$ .
The expression $x^2\ln x$ is not defined at zero but tends to $0$ there, so it has a continuous extension to $[0,1]$ . To evaluate its integral, apply integration by parts first on $[\varepsilon,1]$ , with $f(x)=\ln x$ and $g(x)=x^3/3$ $x^{3} /3$ :

\int_\varepsilon^1x^2\ln x\,\mathrm{d}x=\left[\frac{x^3\ln x}{3}\right]_\varepsilon^1 -\frac{1}{3}\int_\varepsilon^1x^2\,\mathrm{d}x.

Since $\varepsilon^3\ln\varepsilon\to0$ as $\varepsilon\downarrow0$ $0$ , the integral equals $-1/9$ .

Integrating by parts twice and solving for the integral

Compute $\displaystyle I=\int_0^{\pi}\mathrm{e}^x\sin x\,\mathrm{d}x$ $\int_{0}^{π} e^{x} sin x d x$ .

First integration by parts

Take $f(x)=\sin x$ and $g'(x)=\mathrm{e}^x$ $e^{x}$ , so $f'(x)=\cos x$ and $g(x)=\mathrm{e}^x$ $e^{x}$ :

I=\big[\mathrm{e}^x\sin x\big]_0^{\pi}-\int_0^{\pi}\mathrm{e}^x\cos x\,\mathrm{d}x=0-J,

where $J=\int_0^{\pi}\mathrm{e}^x\cos x\,\mathrm{d}x$ $\int_{0}^{π} e^{x} cos x d x$ , because $\sin\pi=\sin0=0$ $0$ .

Second integration by parts, same choice of

g'

Now take $f(x)=\cos x$ and again $g'(x)=\mathrm{e}^x$ $e^{x}$ :

J=\big[\mathrm{e}^x\cos x\big]_0^{\pi}+\int_0^{\pi}\mathrm{e}^x\sin x\,\mathrm{d}x=-\mathrm{e}^{\pi}-1+I.

Solve for

I

Substituting $J$ into $I=-J$ gives $I=\mathrm{e}^{\pi}+1-I$ $I$ , so

I=\frac{\mathrm{e}^{\pi}+1}{2}.

Takeaway. When integrating by parts twice brings back the integral you started with, treat it as an unknown and solve the equation. Keep $g'=\mathrm{e}^x$ in both rounds; switching to $g'=\cos x$ the second time undoes the first step.

Prerequisites¶

Sources¶

Math I: Calculus · Lecture 12

Aliases: IBP, partial integration, integrate by parts Symbols: epsilon, gamma, integral, pi

Last updated: 27 October 2026 11:52