Comprehensive Guide to Integration by Parts and the LI8 Method

Derivation of the Integration by Parts Formula

Integration by parts is a technique derived directly from the product rule of differentiation. The product rule states that the derivative of the product of two functions, uu and vv, is given by:

ddx(uv)=uv+uv\frac{d}{dx}(uv) = u'v + uv'

To derive the integration by parts formula, integrate both sides of this equation with respect to xx:

ddx(uv)dx=uvdx+uvdx\int \frac{d}{dx}(uv) dx = \int u'v dx + \int uv' dx

Because integration and differentiation are inverse processes, the integral of the derivative of (uv)(uv) simply yields the original product (uv)(uv). Thus, the equation becomes:

uv=uvdx+uvdxuv = \int u'v dx + \int uv' dx

Rearranging this expression to solve for one of the integrals results in the standard formula for integration by parts:

uvdx=uvuvdx\int uv' dx = uv - \int u'v dx

In alternative notation commonly used in differential form, this is expressed as:

udv=uvvdu\int u \, dv = uv - \int v \, du

The General Philosophy and Rule of Thumb

The primary purpose of integration by parts is to transform an integral that is difficult or impossible to solve directly into a new integral that is simpler. The method essentially allows for the integration of one component of the integrand while the other component is differentiated.

The Core Rule of Thumb: Integration by parts should be used to integrate the hardest possible chunk of the integrand that is still integrable. By pulling out the most complex part that can be successfully integrated, the remaining integral (the new integral generated by the formula) will hopefully be simple enough to evaluate using standard techniques.

The LI8 Selection Acronym

A critical step in integration by parts is correctly identifying which part of the integrand should serve as uu (to be differentiated) and which should serve as dvdv (to be integrated). The acronym LI8 (sometimes referred to as LIATE) provides a hierarchy for choosing uu. The first function type encountered in the list below should be designated as uu:

  1. L (Logs): Logarithmic functions such as ln(x)\ln(x).

  2. I (Inverse Trig): Inverse trigonometric functions such as arcsin(x)\arcsin(x) or arctan(x)\arctan(x).

  3. 8 (Algebraic): Algebraic expressions including polynomials (xnx^n), roots (x\sqrt{x}), and rational functions.

  4. T (Trig): Standard trigonometric functions such as sin(x)\sin(x) or cos(x)\cos(x).

  5. E (Exponential): Exponential functions such as exe^x.

Once uu is selected based on this order, everything else remaining in the integrand must be designated as dvdv. This hierarchy works because it prioritizes functions that simplify significantly when differentiated (LL, II, AA) and leaves functions that are easy to integrate (TT, EE) to be part of dvdv.

The Box Method Strategy

The box method is a visual organizational tool used to perform integration by parts without strictly relying on the algebraic formula. It utilizes a 2×22 \times 2 grid to organize the components:

  • Upper-Left Corner: Place the chosen uu.

  • Bottom-Right Corner: Place the chosen dvdv (or vv').

  • Vertical Movement (Downwards): Differentiate the value in the top-left to find dudu (or uu'). (Memory aid: DD for Down and Derivative).

  • Vertical Movement (Upwards): Integrate the value in the bottom-right to find vv.

The Rule of Seven: To construct the solution from the box:

  1. Multiply the terms on the diagonal (top-left to bottom-right, resulting in uvuv).

  2. Multiply the terms on the bottom row (bottom-left to bottom-right, which is vduv \, du).

  3. The final result follows the structure: Diagonal Product(Bottom Row Product)\text{Diagonal Product} - \int (\text{Bottom Row Product}).

Procedural Examples and Application

Example 1: xsin(x)dx\int x \sin(x) dx
  • Identify uu: Following LI8, xx is Algebraic and sin(x)\sin(x) is Trig. Therefore, u=xu = x.

  • Identify dvdv: The remainder of the integrand is dv=sin(x)dv = \sin(x).

  • Fill the Box:

    • u=xu = x

    • du=1du = 1

    • dv=sin(x)dv = \sin(x)

    • v=cos(x)v = -\cos(x)

  • Apply the Formula:     (x)(cos(x))(cos(x))(1)dx(x)(-\cos(x)) - \int (-\cos(x))(1) dx     xcos(x)+cos(x)dx-x \cos(x) + \int \cos(x) dx     xcos(x)+sin(x)+C-x \cos(x) + \sin(x) + C

Example 2: arctan(x)dx\int \arctan(x) dx
  • Identify uu: Inverse Trig is higher on the LI8 list, so u=arctan(x)u = \arctan(x).

  • Identify dvdv: Since the entire function is taken as uu, the remaining part is dv=1dv = 1.

  • Fill the Box:

    • u=arctan(x)u = \arctan(x)

    • du=11+x2du = \frac{1}{1 + x^2}

    • v=xv = x

  • Apply the Formula:     xarctan(x)x1+x2dxx \arctan(x) - \int \frac{x}{1 + x^2} dx

  • Evaluate the New Integral: Using U-substitution where w=1+x2w = 1 + x^2 and dw=2xdxdw = 2x \, dx:     xarctan(x)12ln1+x2+Cx \arctan(x) - \frac{1}{2} \ln|1 + x^2| + C

Example 3: xsec(x)tan(x)dx\int x \sec(x) \tan(x) dx
  • Identify uu: Algebraic function xx is chosen as uu.

  • Identify dvdv: Trig part sec(x)tan(x)\sec(x) \tan(x) is chosen as dvdv.

  • Fill the Box:

    • u=xu = x

    • du=1du = 1

    • v=sec(x)v = \sec(x) (since the derivative of sec(x)\sec(x) is sec(x)tan(x)\sec(x)\tan(x)).

  • Apply the Formula:     xsec(x)sec(x)dxx \sec(x) - \int \sec(x) dx     xsec(x)lnsec(x)+tan(x)+Cx \sec(x) - \ln|\sec(x) + \tan(x)| + C

Subtle Errors and Failure Cases

While the LI8 method is highly effective (estimated success rate of 96%96\% to 98%98\%), it can fail or require adjustments in specific scenarios.

Overzealous Collection: In some cases, the LI8 rule may suggest a choice for dvdv that is impossible to integrate in its given form. For example, in x3ex2dx\int x^3 e^{x^2} dx, LI8 suggests u=x3u = x^3 and dv=ex2dv = e^{x^2}. However, ex2e^{x^2} has no elementary antiderivative and is impossible to integrate as is.

  • The Fix: Borrow factors from the algebraic term. Instead of dv=ex2dv = e^{x^2}, use dv=xex2dv = x e^{x^2}, which is integrable via U-substitution (w=x2w = x^2). In this case, uu becomes x2x^2.

Total Failure of LI8: In approximately 0.01%0.01\% of cases, the LI8 hierarchy is completely wrong. This typically happens with integrands like xex(1+x)2\frac{x e^x}{(1+x)^2}.

  • If you choose uu as the algebraic part according to LI8, the resulting integrals become increasingly complex with higher powers in the denominator.

  • To solve this, return to first principles: integrate the hardest possible chunk. The hardest integrable part here is 1(1+x)2dx\int \frac{1}{(1+x)^2} dx, or better yet, using a complex dvdv such as xexx e^x could lead to a solution, though it requires specific insight.

The Problem of Repetitive Integration by Parts

Certain integrals require the application of integration by parts multiple times. For an integral like xnexdx\int x^n e^x dx, you generally must perform integration by parts nn times (the number of times corresponds to the power of the polynomial).

Example: x2exdx\int x^2 e^x dx

  1. First Pass: u=x2u = x^2, dv=exdv = e^x. Result: x2ex2xexdxx^2 e^x - \int 2x e^x dx.

  2. Second Pass: Apply integration by parts again to 2xexdx\int 2x e^x dx with u=2xu = 2x and dv=exdv = e^x. Result: 2xex2exdx2x e^x - \int 2 e^x dx.

  3. Final Combination: Distribute negatives and combine terms to reach the final answer plus a constant CC. This process is labor-intensive and error-prone due to repeated substitution and sign tracking.