Integration by Parts

By the end of this lesson, you will be able to apply integration by parts to evaluate integrals, select $u$ and $dv$ effectively using LIATE, and handle repeated applications.

The Formula

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

This is the "undo" of the product rule: $d(uv) = u, dv + v, du$ → $u, dv = d(uv) - v, du$ → integrate.

LIATE Priority for choosing $u$: Logarithms, Inverse trig, Algebraic (polynomials), Trig, Exponential

Choose $u$ as the first type in LIATE that appears in the integrand. The remaining factor is $dv$.

Examples

Example 1: $\int x e^x, dx$

$u = x$ (Algebraic), $dv = e^x, dx$; $du = dx$, $v = e^x$

$= xe^x - \int e^x, dx = xe^x - e^x + C = e^x(x-1)+C$

Example 2: $\int x^2\ln x, dx$

$u = \ln x$ (Logarithmic — L comes before A in LIATE), $dv = x^2, dx$; $du = \frac{dx}{x}$, $v = \frac{x^3}{3}$

$= \frac{x^3}{3}\ln x - \int \frac{x^3}{3}\cdot\frac{1}{x}, dx = \frac{x^3\ln x}{3} - \frac{1}{3}\int x^2, dx = \frac{x^3\ln x}{3} - \frac{x^3}{9} + C$

Example 3 (Repeated IBP): $\int x^2 e^x, dx$

$u = x^2$, $dv = e^x, dx$ → $xe^x - \int xe^x, dx$

Apply IBP again to $\int xe^x, dx$: $= xe^x - e^x$

Full result: $x^2e^x - 2(xe^x - e^x) + C = e^x(x^2 - 2x + 2) + C$

Example 4 (Circular): $\int e^x\sin x, dx$

Let $I = \int e^x\sin x, dx$

IBP twice → $I = e^x\sin x - e^x\cos x - I$

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

Tabular Method (for Repeated IBP)

For $\int p(x)e^{ax}, dx$ where $p$ is a polynomial, build a table:

Sign	Differentiate ($u$)	Integrate ($dv$)
+	$x^2$	$e^x$
−	$2x$	$e^x$
+	$2$	$e^x$
−	$0$	$e^x$

Multiply along diagonals and alternate signs: