What is the Integration of UV Formula (Integration by Parts)?

The formula for integration of the product of two functions u and v is:

∫u·v dx = u·∫v dx − ∫[du/dx · ∫v dx] dx

Using compact notation with differentiation and integration symbols: ∫u dv = u·v − ∫v du

Where:

• u is the function chosen for differentiation
• v dx (or dv) is the function chosen for integration
• du/dx is the derivative of u
• ∫v dx is the integral of v

This formula converts the problem of integrating u·v into a (hopefully simpler) integral on the right side.

The formula is derived from the product rule of differentiation.

Product rule: d/dx[u·v] = u·(dv/dx) + v·(du/dx)

Rearranging: u·(dv/dx) = d/dx[u·v] − v·(du/dx)

Integrating both sides with respect to x: ∫u·(dv/dx) dx = ∫d/dx[u·v] dx − ∫v·(du/dx) dx

∫u dv = u·v − ∫v du

This is exactly the integration by parts formula. It is a direct consequence of the product rule, which is why both rules are closely paired in calculus courses.

The ILATE rule (also written as LIATE) helps you decide which function to take as u (to differentiate) and which as v (to integrate):

I — Inverse trigonometric functions (sin⁻¹x, cos⁻¹x, tan⁻¹x) L — Logarithmic functions (log x, ln x) A — Algebraic functions (polynomials: xⁿ, x², x, constants) T — Trigonometric functions (sin x, cos x, tan x) E — Exponential functions (eˣ, aˣ)

The function that appears EARLIER in ILATE is taken as u.

Why this order? Functions higher in the list become simpler when differentiated (e.g., log x becomes 1/x), while functions lower in the list are easy to integrate (e.g., eˣ integrates to itself).

Example: In ∫x·eˣ dx → x is algebraic (A), eˣ is exponential (E). A comes before E in ILATE, so u = x, dv = eˣ dx.

Example 1: ∫x·eˣ dx Choose u = x, dv = eˣ dx du = dx, v = eˣ Apply formula: ∫x·eˣ dx = x·eˣ − ∫eˣ dx = x·eˣ − eˣ + C = eˣ(x − 1) + C

Example 2: ∫x·sin x dx Choose u = x (A), dv = sin x dx (T) du = dx, v = −cos x ∫x·sin x dx = x·(−cos x) − ∫(−cos x) dx = −x cos x + sin x + C

Example 3: ∫ln x dx Write as ∫ln x · 1 dx; u = ln x (L), dv = dx (A) du = 1/x dx, v = x ∫ln x dx = x·ln x − ∫x · (1/x) dx = x ln x − ∫1 dx = x ln x − x + C

Example 4: ∫x²·eˣ dx (apply integration by parts twice) First pass: u = x², dv = eˣ dx → du = 2x dx, v = eˣ = x²·eˣ − ∫2x·eˣ dx Second pass on ∫2x·eˣ dx: u = 2x, dv = eˣ dx → = 2x·eˣ − 2eˣ Final answer: x²eˣ − 2xeˣ + 2eˣ + C = eˣ(x² − 2x + 2) + C

Example 5: ∫eˣ·sin x dx (cyclic case — apply by parts twice) Let I = ∫eˣ sin x dx Pass 1: u = sin x, dv = eˣ dx → I = eˣ sin x − ∫eˣ cos x dx Pass 2 on ∫eˣ cos x dx: u = cos x, dv = eˣ dx → = eˣ cos x + ∫eˣ sin x dx = eˣ cos x + I So: I = eˣ sin x − (eˣ cos x + I) 2I = eˣ sin x − eˣ cos x I = eˣ(sin x − cos x)/2 + C

1. Cyclic integrals: When integrating by parts twice brings you back to the original integral I, rearrange algebraically to solve for I.

2. Single function integrals: Integrals like ∫ln x dx and ∫sin⁻¹x dx can be solved by writing them as products with 1 (e.g., ∫ln x · 1 dx) and applying integration by parts.

3. Tabular method (for repeated application): Useful when a polynomial multiplies a function that integrates repeatedly. Create two columns — derivatives of u (until zero) and integrals of dv — then multiply diagonally with alternating signs.

4. When NOT to use integration by parts: If substitution (u-substitution) works more easily, prefer it. Integration by parts is best when the integrand is a product of functions from different categories in ILATE.

5. Common results derived via integration by parts:

• ∫x eˣ dx = eˣ(x−1) + C
• ∫x sin x dx = sin x − x cos x + C
• ∫x cos x dx = cos x + x sin x + C
• ∫ln x dx = x ln x − x + C
• ∫x² eˣ dx = eˣ(x²−2x+2) + C