Integration by Parts

Part 1 of 7 — The Formula

Integration by Parts Formula

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

LIATE Rule for Choosing $u$

Choose $u$ in this priority order:

1. Logarithmic ($ln x$, $log x$)
2. Inverse trig ($arctan x$, $arcsin x$)
3. Algebraic ($x^2$, $x$, polynomials)
4. Trigonometric ($sin x$, $cos x$)
5. Exponential ($e^x$, $2^x$)

Worked Example

$int x,e^x,dx$

$u = x$, $dv = e^x,dx$

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

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

Integration by Parts 🎯

Key Takeaways — Part 1

1. $\int u\,dv = uv - \int v\,du$
2. Use LIATE to choose $u$
3. $\int \ln x\,dx = x\ln x - x + C$

Integration by Parts

Part 2 of 7 — Tabular Method

The Tabular (Column) Method

For $int ( ext{polynomial}) cdot ( ext{easy to integrate}),dx$:

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

Multiply diagonally and alternate signs!

Tabular Method 🎯

Key Takeaways — Part 2

1. Tabular method is fast for polynomial × exponential/trig
2. Alternate signs: $+, -, +, -, ...$
3. Multiply diagonally across columns

Integration by Parts

Part 3 of 7 — Cycling (Boomerang) Problems

When IBP Cycles Back

For $int e^x sin x,dx$, doing IBP twice brings back the original integral!

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

IBP #1: $u = sin x$, $dv = e^x,dx$ $I = e^x sin x - int e^x cos x,dx$

IBP #2: $u = cos x$, $dv = e^x,dx$ $I = e^x sin x - e^x cos x - int e^x(-sin x),dx$ $I = e^x sin x - e^x cos x - I$

$2I = e^x(sin x - cos x)$

I = rac{e^x(sin x - cos x)}{2} + C

Cycling IBP 🎯

Key Takeaways — Part 3

1. $e^x \cdot \text{trig}$ always cycles after 2 applications
2. Solve for $I$ algebraically
3. Keep the same choice pattern for $u$ both times!

Integration by Parts

Part 4 of 7 — Definite Integrals with IBP

Formula for Definite Integrals

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

Worked Example

$int_0^1 xe^x,dx = [xe^x]_0^1 - int_0^1 e^x,dx = e - [e^x]_0^1 = e - (e-1) = 1$

Definite IBP 🎯

Key Takeaways — Part 4

1. Apply bounds to the $uv$ term
2. Bounds carry through to the remaining integral

Integration by Parts

Part 5 of 7 — Special Cases

Inverse Trig Integrals

$int arctan x,dx$: $u = arctan x$, $dv = dx$

= xarctan x - int rac{x}{1+x^2},dx = xarctan x - rac{1}{2}ln(1+x^2) + C

$int ln x$ derivatives

$int (ln x)^2,dx$: $u = (ln x)^2$, $dv = dx$

$= x(ln x)^2 - 2int ln x,dx = x(ln x)^2 - 2(xln x - x) + C$

Special Cases 🎯

Key Takeaways — Part 5

1. Inverse trig and logarithmic functions always go as $u$
2. The resulting integral usually becomes a u-substitution

Integration by Parts

Part 6 of 7 — Practice Workshop

Mixed IBP Practice 🎯

Workshop Complete!

Integration by Parts — Review

Part 7 of 7 — Final Assessment

Final Assessment 🎯

Integration by Parts — Complete! ✅