Integration by Parts – AP Calculus

Integration by Parts is the counterpart to the Product Rule for differentiation. It's used when the integrand is a product of two functions that don't play nicely with substitution. The formula transforms one integral into another that (hopefully) is easier to evaluate.

🎯 In this section you will learn

The Integration by Parts formula: $\int u \, dv = uv - \int v \, du$
How to choose $u$ and $dv$ (LIATE rule)
Applying integration by parts once
Repeated integration by parts (tabular method)
Integration by parts for definite integrals

📌 The Formula

$$ \int u \, dv = uv - \int v \, du $$ Let $u$ and $dv$ be parts of the integrand. Then $du$ and $v$ are derived from them.

💡 How to Choose $u$ and $dv$ (LIATE Rule)

Choose $u$ in this order: Logarithmic, Inverse trigonometric, Algebraic (polynomial), Trigonometric, Exponential.
The function that comes first in LIATE becomes $u$; the rest becomes $dv$.

Example 1Integration by Parts: $x e^x$

Find $\displaystyle \int x e^x \, dx$.

①

Choose $u$ and $dv$

$u = x$ (algebraic), $dv = e^x dx$ (exponential)

②

Find $du$ and $v$

$du = dx$, $v = \int e^x dx = e^x$

③

Apply formula

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

Example 2Integration by Parts: $x \ln x$

Find $\displaystyle \int x \ln x \, dx$.

①

Choose $u$ and $dv$

$u = \ln x$ (logarithmic), $dv = x dx$ (algebraic)

②

Find $du$ and $v$

$du = \frac{1}{x} dx$, $v = \frac{x^2}{2}$

③

Apply formula

$\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$

Example 3Integration by Parts: $x \cos x$

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

$u = x$, $dv = \cos x dx$
$du = dx$, $v = \sin x$
$\int x \cos x dx = x \sin x - \int \sin x dx = x \sin x + \cos x + C$

📌 Repeated Integration by Parts (Tabular Method)

Sometimes you need to apply integration by parts more than once. The tabular method is a systematic way to handle repeated integration by parts.

Example 4Repeated Integration by Parts

Find $\displaystyle \int x^2 e^x \, dx$.

①

First application

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

②

Second application

$\int 2x e^x dx = 2(x e^x - e^x) + C = 2x e^x - 2e^x + C$

③

Combine

$\int x^2 e^x dx = x^2 e^x - 2x e^x + 2e^x + C = e^x(x^2 - 2x + 2) + C$

📌 Integration by Parts for Definite Integrals

\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du

Example 5Definite Integral by Parts

Find $\displaystyle \int_0^1 x e^x \, dx$.

Using Example 1: $\int x e^x dx = e^x(x - 1)$
$\int_0^1 x e^x dx = [e^x(x - 1)]_0^1 = (e^1(0)) - (e^0(-1)) = 0 - (-1) = 1$

⚠️ Common Mistakes

Choosing the wrong $u$: If the integral gets more complicated after one application, try swapping $u$ and $dv$.
Forgetting the minus sign: The formula is $uv - \int v du$, not $uv + \int v du$.
Not completing repeated applications: Some integrals require integration by parts multiple times (like $x^2 e^x$).

🔍 Key Takeaways

Integration by Parts: $\int u dv = uv - \int v du$.
Use LIATE to choose $u$ (Log, Inverse trig, Algebraic, Trig, Exponential).
Some integrals require repeated integration by parts.
For definite integrals: $\int_a^b u dv = [uv]_a^b - \int_a^b v du$.

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