Integration by Parts: Key Concepts, Derivation, and Examples

Product Rule Recap and the Idea Behind Integration by Parts

  • Product rule (Calc I) for a product of two functions u(x) and v(x):
    • (uv)′=u′v+uv′(uv)' = u'v + uv'
  • In Calc II, we can (formally) integrate the product rule with respect to x:
    • ∫(uv)′ dx=∫(u′v+uv′) dx=uv+C\int (uv)' \,dx = \int (u'v + uv') \,dx = uv + C
  • Split the left side into two integrals:
    • ∫u′v dx+∫uv′ dx=uv+C\int u'v \,dx + \int uv' \,dx = uv + C
  • Solve for the integral you want (you pick which one to isolate):
    • ∫uv′ dx=uv−∫u′v dx\int uv' \,dx = uv - \int u'v \,dx
  • This is the standard integration by parts formula when we set:
    • dv=v′ dxanddu=u′ dxdv = v'\,dx \quad\text{and}\quad du = u'\,dx
    • so that u≡u, dv≡dvu\equiv u, \ dv\equiv dv and v=∫dv=v, du=duv = \int dv = v, \ du = du
  • Practical takeaway: integration by parts exchanges an integral of a product where one factor is differentiated and the other is integrated, for an integral that is hopefully easier to compute.
  • Note about constants: the antiderivative introduces a constant of integration, so results end with +C+C.

The Integration by Parts Formula (Standard Form)

  • Start with a product of two functions: a part you call u and a part you call dv (the differential piece):
    • Define du=u′dx, dv=v′dx, v=∫dvdu = u' dx, \ dv = v' dx, \ v = \int dv
  • The formula is:
    • ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
  • Key interpretation: you replace the original integral (the red one) with a new one (the purple one) that is hopefully easier to evaluate.
  • End result always includes a final plus constant: +C).+C).

Notation and Generality

  • Everything is a function of x here; the differential could be with respect to any variable (dx, dy, dt, etc.).
  • We often compress the notation by writing simply: ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du, with the understanding that $du$ and $dv$ come from differentiating/integrating with respect to x (or the chosen variable).
  • The choice of the letters (u and dv) is conventional; the important part is which part you differentiate (to get du) and which part you integrate (to get v).
  • You may see or hear about choosing $u$ and $dv$ and then rewriting the integral; the order matters for usefulness, not just correctness.

Why IBP Is Useful

  • It lets you turn an integral you don’t know into one you might know how to do.
  • It is especially powerful for products where one factor becomes simpler when differentiated and the other has a simple antiderivative.
  • It is a tool to connect substitutions, known antiderivatives, and algebraic manipulation to tackle tricky integrals.
  • In practice, we often apply IBP one or a few times; many problems do not require dozens of iterations.

How to Choose u and dv: Guidelines and Pitfalls

  • Core guideline: choose $u$ so that $du$ is easier to integrate or simpler than the original; choose $dv$ so that its antiderivative $v$ is easy to compute.
  • Common heuristic (often taught): pick $u$ to be a function that becomes simpler when differentiated (e.g., polynomials, logarithms, inverse trigonometric functions) and pick $dv$ to be a function whose antiderivative is easy (e.g., exponential, sine, cosine).
  • Important caveat from practice:
    • There is typically only one sensible ordering of $u$ and $dv$ for a given product; choosing the wrong ordering can make the remaining integral harder or even intractable.
    • Example caution: putting $u = \ln x$ often leads to $du = \frac{1}{x} dx$ and an integral $\int v\,du = \int v \frac{1}{x} dx$ that may not be easier; sometimes you end up with a nontrivial integral instead of simplifying.
  • When you have a term whose derivative would make the next integral easy (e.g., derivative becoming constant), that choice is especially favorable (e.g., $u = x$ so that $du = dx$ is constant).
  • Sometimes there is no good way to apply IBP in a given step; you may have to recognize that a different approach or another round of IBP is required later.
  • The sign in the IBP formula is fixed: the subtraction term always appears as a minus sign in ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du; any sign changes come from the algebra, not from the formula itself.

Common Mistakes and Clarifications

  • It is mathematically correct to derive an alternative IBP setup (e.g., swapping roles of $u$ and $dv$). However, not all choices are useful; some produce harder integrals.
  • There is no triple-product IBP formula; you cannot do a single IBP that directly handles three multiplicative factors. Instead, you can choose which two factors form the product at each step and possibly apply IBP multiple times.
  • When you do multiple IBPs in a row, you are effectively reducing the original integral by expressing it in terms of itself with altered coefficients or a simpler integral that can eventually cancel, but you must track the signs and constants carefully.
  • The negative sign in the IBP result originates from moving the integral term to the other side of the equation during derivation; it remains in front of the remaining integral at each step.
  • Some instructors also discuss a Tabular Method (rapid repeated IBP) as a fast algorithmic approach; however, this is less emphasized in newer curricula and is often replaced by computational tools.

Worked Example: (\int x e^x \,dx) (Standard Choice)

  • Choose: u=x,dv=ex dxu = x, \quad dv = e^x\,dx
  • Then: du=dx,v=∫ex dx=exdu = dx, \quad v = \int e^x\,dx = e^x
  • Apply IBP:
    • ∫xex dx=uv−∫v du=xex−∫ex dx\int x e^x \,dx = uv - \int v\,du = x e^x - \int e^x\,dx
    • =xex−ex+C=ex(x−1)+C= x e^x - e^x + C = e^x(x - 1) + C
  • Remark: The remaining integral $\int e^x\,dx$ is straightforward, which is why this ordering is effective.

Alternative Choice Illustration: Why Another Ordering Can Be Worse

  • Try: u=ex,dv=x dxu = e^x, \quad dv = x\,dx
  • Then: du=ex dx,v=∫x dx=x22du = e^x\,dx, \quad v = \int x\,dx = \frac{x^2}{2}
  • IBP gives:
    • ∫xex dx=uv−∫v du=exx22−∫x22ex dx\int x e^x\,dx = uv - \int v\,du = e^x\frac{x^2}{2} - \int \frac{x^2}{2} e^x\,dx
  • This transforms the problem into a more complicated integral: ∫x22ex dx\int \frac{x^2}{2} e^x\,dx which is not immediately easier.
  • Takeaway: While mathematically valid, choice affects usefulness; always aim to reduce the difficulty of the remaining integral.
  • Additional note from the discussion: sometimes you can apply IBP again to the new integral (e.g., for higher powers of x), but you may end up cycling back or needing multiple iterations.

Repeated Applications: When You Might Need It

  • Some integrals require applying IBP more than once:
    • Example pattern: integrals of the form ∫xnex dx\int x^n e^x \ dx may require multiple IBP steps.
  • After one IBP, you may obtain an integral of a similar form but with a lower power of x or with a factor that makes the second IBP effective.
  • In the transcript, the instructor demonstrates applying IBP twice to a polynomial times an exponential and shows how terms propagate and cancel, leading to a closed form.

Tabular Method (Rapid Repeated IBP) and Its Place in Teaching

  • Tabular Method (also called Rapid Repeated IBP) is a technique some students learn to handle repeated IBP quickly.
  • The instructor notes:
    • It accelerates problems requiring many IBP iterations, but with modern computer algebra systems, manual rapid repetition is less essential.
    • The instructor does not teach it, but acknowledges it exists and can be used if a student explores it.
  • Rationale given for not teaching it: computing power and software can perform many IBP steps quickly; the method is considered older and less practical today.

Practical Takeaways for Exams and Practice

  • Always verify what you are differentiating and what you are integrating: choose to differentiate the part that simplifies, and integrate the part with a simple antiderivative.
  • Keep track of the signs: IBP introduces and propagates negative signs; be careful with subtracting the remaining integral.
  • After applying IBP, assess whether the remaining integral is easier; if not, try a different split for u and dv.
  • You can iterate IBP multiple times when needed, but the number of iterations on exams is typically limited.
  • Remember to append the constant of integration at the end: +C+C.
  • If you have a product with a natural log, polynomial, or inverse trig, think about how the derivative reduces the complexity; sometimes ln x as u can complicate the remaining integral if dv is not chosen carefully.
  • In real-world practice, use a computer algebra system for heavy repeated IBP work; know the basic IBP formula and how to apply it by hand for common problems.

Quick Reference: Key Formulas and Concepts

  • Product rule: ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv'
  • Integration by parts (standard form): ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
  • Relationship to product rule: integrating the product rule gives the IBP formula.
  • Common integration targets for choosing u and dv: polynomial, log, inverse trig for u; exponential, sine, cosine for dv.
  • Final form always includes a constant: +C+C.
  • In iterative IBP, you may encounter expressions like ∫xnexdx\int x^n e^x dx that require repeated application.
  • Tabular Method: an algorithmic approach to performing multiple IBPs quickly (not required to know for all courses, and often replaced by computer algebra).

Quick Problem Checklist (for practice)

  • Identify the product structure: is there a clear split into a part to differentiate and a part to integrate?
  • Choose u to simplify upon differentiation and dv to have a simple antiderivative.
  • Compute du and v, apply ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du.
  • If the remaining integral is still nontrivial, consider a new choice for u and dv and repeat (watch for signs).
  • Consider whether a tabular approach would be beneficial (optional).
  • Verify the result by differentiating (optional check).