Integration by Parts
Integration technique from the product rule.
Formula
Variables
Example
∫ x·e^x dx = x·e^x − ∫ e^x dx = e^x(x − 1) + C.
How it works
Integration by parts is the integral version of the product rule for derivatives. Since d(uv) = u\,dv + v\,du, rearranging and integrating both sides gives \int u\,dv = uv - \int v\,du. It is especially useful when one factor becomes simpler after differentiation while the other is easy to integrate.
When to use it
Use this when an integral contains a product of two different types of functions, such as a polynomial with an exponential, logarithm, inverse trig function, or trig function. It is also a standard tool for reducing powers, evaluating definite integrals, and deriving formulas in physics and probability.
Worked example, step by step
Evaluate \int_0^1 x\ln x\,dx Let u = \ln x, so du = \frac{1}{x}dx Let dv = x\,dx, so v = \frac{x^2}{2} Apply integration by parts: \int_0^1 x\ln x\,dx = \left[\frac{x^2}{2}\ln x\right]_0^1 - \int_0^1 \frac{x^2}{2}\cdot \frac{1}{x}\,dx = \left[\frac{x^2}{2}\ln x\right]_0^1 - \frac{1}{2}\int_0^1 x\,dx Now evaluate the boundary term: At x = 1, \frac{1^2}{2}\ln 1 = 0 As x \to 0^+, \frac{x^2}{2}\ln x \to 0 So the boundary term is 0 Then compute the remaining integral: \frac{1}{2}\int_0^1 x\,dx = \frac{1}{2}\left[\frac{x^2}{2}\right]_0^1 = \frac{1}{4} Therefore, \int_0^1 x\ln x\,dx = 0 - \frac{1}{4} = -\frac{1}{4}
Common mistakes to avoid
- Choosing u and dv in a way that makes the new integral harder than the original
- Forgetting the minus sign in \u222b u\,dv = uv - \u222b v\,du
- Dropping boundary terms or evaluating them incorrectly in definite integrals
Did You Know?
Integration by parts is the reverse of the product rule — it trades one integral for an easier one.
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