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Calculus II (Integration & Series) Unit 2: Integration by parts
Integration by parts reverses the product rule. Use it when the integrand is a product of two different kinds of functions.
Covers: Integration by parts, tabular method
Study guide
Integration by parts reverses the product rule: ∫ u dv = uv − ∫ v du. Split the integrand into u (a part that gets simpler when differentiated) and dv (a part you can integrate). Find du and v, apply the formula, and evaluate the new integral.
The LIATE order is a guide for choosing u: Logarithms, Inverse trig, Algebraic (polynomials), Trig, Exponential. Pick u from whichever type comes first. For ∫ x eˣ dx, u = x and dv = eˣ dx, giving x eˣ − eˣ + C. For ∫ ln x dx, take u = ln x and dv = dx, giving x ln x − x + C. The same idea works for ∫ arctan x dx.
Sometimes you must apply parts more than once, such as ∫ x² eˣ dx, where u = x² must be reduced twice. The tabular method speeds this up: list repeated derivatives of u in one column until reaching 0, repeated integrals of dv in another, multiply diagonally, and alternate signs.
Some integrals are cyclic, like ∫ eˣ sin x dx. After applying parts twice, the original integral reappears on the right side. Move it to the left, combine, and solve for it algebraically.
For definite integrals, evaluate the uv term at the limits as well: ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du.
If integration by parts leaves ∫ tan x dx, write tan x = sin x/cos x and let u = cos x to get −ln|cos x| + C = ln|sec x| + C. Integration by parts works even when f is known only from given values. For ∫ₐᵇ x f′(x) dx, take u = x and dv = f′(x) dx to get [x f(x)]ₐᵇ − ∫ₐᵇ f(x) dx, then substitute the given values of f and of its integral.
Some integrals need a substitution before parts. For ∫ e^(√x) dx, let w = √x, so x = w² and dx = 2w dw, then integrate 2w eʷ by parts. For ∫ (ln x)ⁿ dx, take u = (ln x)ⁿ and dv = dx. Each application lowers the power of the logarithm by one until only ∫ dx remains.
Key ideas
- Pick u to be the factor that gets simpler when differentiated.
- LIATE is a rule of thumb for u: Log, Inverse trig, Algebraic, Trig, Exponential.
- Sometimes you need to apply parts twice, or solve for the original integral when it reappears.
- Tabular integration speeds up cases like xⁿ·eˣ where u eventually differentiates to 0.
Formulas and rules
Common mistakes
- Picking u = eˣ and making the integral harder.
- Dropping the minus sign in uv − ∫ v du.
Words to know
- Integration by parts formula
- ∫ u dv = uv − ∫ v du
- LIATE
- Logs, Inverse trig, Algebraic, Trig, Exponential: order for choosing u in parts.
- Tabular integration
- Repeatedly differentiate u and integrate dv in columns; handy when u is a polynomial.
- ∫ ln x dx
- x ln x − x + C
- Cyclic integration by parts
- When the integral reappears, solve for it algebraically.
- Choosing u (parts)
- A part that gets simpler when differentiated (LIATE order).
- ∫ x·eˣ dx
- x·eˣ − eˣ + C
- Repeated integration by parts
- Apply parts again until the integral is simple, as with ∫ x²eˣ dx.
- Parts for definite integrals
- ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du
Sample practice questions
19 more questions like these, from level 1 to 5, are in the practice set.
For ∫ x sin x dx, the best choice of u is…
- sin x
- x sin x
- cos x
- x
Show the answer and steps
Answer: D) x
- LIATE: algebraic (x) comes before trig.
- x gets simpler when differentiated (to 1).
- u = x.
∫ x eˣ dx
- (x − 1)eˣ + C
- x eˣ + C
- x²eˣ/2 + C
- (x + 1)eˣ + C
Show the answer and steps
Answer: A) (x − 1)eˣ + C
- u = x, dv = eˣ dx → du = dx, v = eˣ.
- ∫ = x eˣ − ∫ eˣ dx = x eˣ − eˣ.
- (x − 1)eˣ + C.
∫ x cos x dx
- sin x + C
- −x sin x + cos x + C
- x sin x + cos x + C
- x sin x − cos x + C
Show the answer and steps
Answer: C) x sin x + cos x + C
- u = x, dv = cos x dx → du = dx, v = sin x.
- ∫ = x sin x − ∫ sin x dx.
- = x sin x + cos x + C.