Comprehensive Study Notes on Integration by Parts

Overview and Fundamental Concepts

  • Integration by Parts is a core integration technique introduced in calculus (specifically Section 3.1 of standard textbooks). It represents the first major non-review technique following uu-substitution.
  • While uu-substitution is fundamentally derived from rolling back the Chain Rule for composite functions, Integration by Parts is fundamentally derived from rolling back the Product Rule for differentiation.
  • The underlying principle connecting differentiation and integration is that taking the antiderivative of a derivative recovers the original function, modulo an arbitrary integration constant (+C+ C).

Derivation of the Integration by Parts Formula

  • Consider two differentiable functions defined as u(x)u(x) and v(x)v(x).
  • Applying the Product Rule for differentiation yields:

ddx[u(x)v(x)]=u(x)v′(x)+u′(x)v(x)\frac{d}{dx}[u(x) v(x)] = u(x) v'(x) + u'(x) v(x)

  • Expressed in differential form, this is:

d(uv)=u dv+v dud(u v) = u\,dv + v\,du

  • Integrating both sides of the equation with respect to xx gives:

∫ddx[u(x)v(x)] dx=∫u(x)v′(x) dx+∫v(x)u′(x) dx\int \frac{d}{dx}[u(x) v(x)]\,dx = \int u(x) v'(x)\,dx + \int v(x) u'(x)\,dx

uv=∫u dv+∫v duu v = \int u\,dv + \int v\,du

  • Rearranging terms to isolate ∫u dv\int u\,dv produces the standard Integration by Parts formula:

∫u dv=uv−∫v du\int u\,dv = u v - \int v\,du

  • The formula connects four distinct components: uu, vv, dudu, and dvdv. Always remember to include the constant of integration +C+ C in the final indefinite integral evaluation.

Standard Procedure for Integration by Parts

  • Step 1: Recognition

    • Identify that the integrand consists of two functions multiplied together.
    • Verify that standard uu-substitution is not applicable (i.e., there is no clear inner function whose derivative is multiplying the outside).

  • Step 2: Make a Choice (uu and dvdv)

    • Select one portion of the integrand to be uu and the remaining portion (which must include the differential dxdx or dtdt) to be dvdv.
    • Ensure every term in the original integral is assigned to either uu or dvdv.

  • Step 3: Construct the Evaluation Table

    • Set up a 2×22 \times 2 grid:

u=[Chosen Function]  ⟹  du=[Derivative of u]   dxu = \text{[Chosen Function]} \quad \implies \quad du = \text{[Derivative of } u \text{] \, dx}

dv=[Chosen Differential]  ⟹  v=[Antiderivative of dv]dv = \text{[Chosen Differential]} \quad \implies \quad v = \text{[Antiderivative of } dv \text{]}

  • Step 4: Substitute into Formula

    • Plug uu, vv, dudu, and dvdv into ∫u dv=uv−∫v du\int u\,dv = u v - \int v\,du.

  • Step 5: Evaluate the Remaining Integral

    • Compute ∫v du\int v\,du. This new integral should be simpler than the original integral.
    • Add the constant of integration +C+ C.

Detailed Examples and Applications

  • Example 1: Polynomial and Trigonometric Product

∫xcos⁡(x) dx\int x \cos(x)\,dx

  • Step 1 (Choice): Let u=xu = x and dv=cos⁡(x) dxdv = \cos(x)\,dx.
  • Step 2 (Table):

u=x  ⟹  du=1 dx=dxu = x \implies du = 1\,dx = dx

dv=cos⁡(x) dx  ⟹  v=sin⁡(x)dv = \cos(x)\,dx \implies v = \sin(x)

  • Step 3 (Apply Formula):

∫xcos⁡(x) dx=xsin⁡(x)−∫sin⁡(x) dx\int x \cos(x)\,dx = x \sin(x) - \int \sin(x)\,dx

  • Step 4 (Evaluate): Since the antiderivative of sin⁡(x)\sin(x) is −cos⁡(x)-\cos(x):

∫xcos⁡(x) dx=xsin⁡(x)−(−cos⁡(x))+C=xsin⁡(x)+cos⁡(x)+C\int x \cos(x)\,dx = x \sin(x) - (-\cos(x)) + C = x \sin(x) + \cos(x) + C

  • Verification via Derivative:

ddx[xsin⁡(x)+cos⁡(x)]=(1⋅sin⁡(x)+xcos⁡(x))−sin⁡(x)=xcos⁡(x)\frac{d}{dx}[x \sin(x) + \cos(x)] = (1 \cdot \sin(x) + x \cos(x)) - \sin(x) = x \cos(x)

  • Example 2: Polynomial and Exponential Product

∫xex dx\int x e^x\,dx

  • Step 1 (Choice): Let u=xu = x and dv=ex dxdv = e^x\,dx.
  • Step 2 (Table):

u=x  ⟹  du=dxu = x \implies du = dx

dv=ex dx  ⟹  v=exdv = e^x\,dx \implies v = e^x

  • Step 3 (Apply Formula & Evaluate):

∫xex dx=xex−∫ex dx=xex−ex+C\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C

  • Verification via Derivative:

ddx[xex−ex+C]=(ex+xex)−ex=xex\frac{d}{dx}[x e^x - e^x + C] = (e^x + x e^x) - e^x = x e^x

  • Example 3: Comparison with Algebraic Expansion

∫t(2−t) dt\int t(2 - t)\,dt

  • Method A (Direct Expansion - Preferred/Simpler):

∫(2t−t2) dt=t2−t33+C\int (2t - t^2)\,dt = t^2 - \frac{t^3}{3} + C

  • Method B (Integration by Parts - For Demonstration):
    • Let u=t  ⟹  du=dtu = t \implies du = dt.
    • Let dv=(2−t) dt  ⟹  v=2t−t22dv = (2 - t)\,dt \implies v = 2t - \frac{t^2}{2}.
    • Apply formula:

∫t(2−t) dt=t(2t−t22)−∫(2t−t22) dt\int t(2 - t)\,dt = t \left(2t - \frac{t^2}{2}\right) - \int \left(2t - \frac{t^2}{2}\right)\,dt

=2t2−t32−(t2−t36)+C= 2t^2 - \frac{t^3}{2} - \left(t^2 - \frac{t^3}{6}\right) + C

=2t2−t32−t2+t36+C=t2−3t36+t36+C=t2−t33+C= 2t^2 - \frac{t^3}{2} - t^2 + \frac{t^3}{6} + C = t^2 - \frac{3t^3}{6} + \frac{t^3}{6} + C = t^2 - \frac{t^3}{3} + C

  • Example 4: Negative Exponential / Decay Functions

∫xe−x dx\int x e^{-x}\,dx

  • Context: Exponential decay functions like e−xe^{-x} are ubiquitous across advanced STEM disciplines (comprising approximately 83\frac{8}{3} of physics topics across courses 240, 250, 251, 252, 253, 254, 255, and 256, as well as 113\frac{11}{3} of differential equations curriculum).
  • Antiderivative Note: The antiderivative of e−xe^{-x} is −e−x-e^{-x}.
  • Step 1 (Choice): Let u=xu = x and dv=e−x dxdv = e^{-x}\,dx.
  • Step 2 (Table):

u=x  ⟹  du=dxu = x \implies du = dx

dv=e−x dx  ⟹  v=−e−xdv = e^{-x}\,dx \implies v = -e^{-x}

  • Step 3 (Apply Formula & Evaluate):

∫xe−x dx=x(−e−x)−∫(−e−x) dx\int x e^{-x}\,dx = x(-e^{-x}) - \int (-e^{-x})\,dx

=−xe−x+∫e−x dx= -x e^{-x} + \int e^{-x}\,dx

=−xe−x−e−x+C= -x e^{-x} - e^{-x} + C

  • Example 5: Logarithmic Functions

∫x2ln⁡(x) dx\int x^2 \ln(x)\,dx

  • Step 1 (Choice): Attempting u=x2u = x^2 requires integrating ln⁡(x)\ln(x) for dvdv, which is complex. Therefore, invert choices:
    • Let u=ln⁡(x)u = \ln(x) and dv=x2 dxdv = x^2\,dx.
  • Step 2 (Table):

u=ln⁡(x)  ⟹  du=1x dxu = \ln(x) \implies du = \frac{1}{x}\,dx

dv=x2 dx  ⟹  v=x33dv = x^2\,dx \implies v = \frac{x^3}{3}

  • Step 3 (Apply Formula & Evaluate):

∫x2ln⁡(x) dx=(x33)ln⁡(x)−∫(x33)(1x) dx\int x^2 \ln(x)\,dx = \left(\frac{x^3}{3}\right) \ln(x) - \int \left(\frac{x^3}{3}\right) \left(\frac{1}{x}\right)\,dx

=x33ln⁡(x)−13∫x2 dx= \frac{x^3}{3} \ln(x) - \frac{1}{3} \int x^2\,dx

=x33ln⁡(x)−x39+C= \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C

Strategies for Selecting uu and dvdv (The LIATE Rule)

  • Selection Hierarchy (LIATE):
    • When deciding which function to assign to uu, choose the function type that appears highest on the following priority list:
    1. L - Logarithmic Functions (e.g., ln⁡(x)\ln(x))
    2. I - Inverse Trigonometric Functions (e.g., arctan⁡(x)\arctan(x))
    3. A - Algebraic Functions / Polynomials (e.g., xx, x2x^2, tt)
    4. T - Trigonometric Functions (e.g., sin⁡(x)\sin(x), cos⁡(x)\cos(x))
    5. E - Exponential Functions (e.g., exe^x, e−xe^{-x})
  • General Objective:
    • Select uu such that dudu becomes simpler (or no more complex) than uu.
    • Select dvdv such that its antiderivative vv can be easily computed.
  • Consequences of an Incorrect Choice:
    • Consider attempting ∫xex dx\int x e^x\,dx with u=exu = e^x and dv=x dxdv = x\,dx:

u=ex  ⟹  du=ex dxu = e^x \implies du = e^x\,dx

dv=x dx  ⟹  v=x22dv = x\,dx \implies v = \frac{x^2}{2}

  • Substituting into the formula gives:

∫xex dx=x22ex−∫x22ex dx\int x e^x\,dx = \frac{x^2}{2} e^x - \int \frac{x^2}{2} e^x\,dx

  • The new integral ∫x22ex dx\int \frac{x^2}{2} e^x\,dx is more complex than the original integral ∫xex dx\int x e^x\,dx. An increase in complexity is a clear signal to abort the current path, swap the choices of uu and dvdv, and restart.

Repeated Integration by Parts (Higher-Order Polynomials)

  • When the algebraic term uu is a higher power (e.g., x2x^2, x3x^3), a single application of integration by parts reduces the exponent by 11, requiring repeated iterations of the technique.
  • Example: Evaluating ∫x2ex dx\int x^2 e^x\,dx
    • First Application:
    • Let u=x2  ⟹  du=2x dxu = x^2 \implies du = 2x\,dx
    • Let dv=ex dx  ⟹  v=exdv = e^x\,dx \implies v = e^x
    • Result:

∫x2ex dx=x2ex−∫2xex dx\int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx

  • Second Application:
    • The remaining integral ∫2xex dx=2∫xex dx\int 2x e^x\,dx = 2 \int x e^x\,dx still contains a product of an algebraic term and an exponential term.
    • Perform integration by parts a second time on ∫xex dx\int x e^x\,dx:
      • Let u=x  ⟹  du=dxu = x \implies du = dx
      • Let dv=ex dx  ⟹  v=exdv = e^x\,dx \implies v = e^x
      • Result for inner integral: xex−exx e^x - e^x
  • Final Combination:

∫x2ex dx=x2ex−2(xex−ex)+C=x2ex−2xex+2ex+C\int x^2 e^x\,dx = x^2 e^x - 2(x e^x - e^x) + C = x^2 e^x - 2x e^x + 2e^x + C

  • General Rule: An initial power of xnx^n multiplied by an exponential or trigonometric function requires applying integration by parts nn times.

Questions & Discussion

  • Question: When choosing u=x2u = x^2 for ∫x2ex dx\int x^2 e^x\,dx, does the expression obtained at the end of the first step mean the problem requires doing integration by parts a second time?

  • Response: Yes. After the first application, the expression contains an integral with x1x^1 (specifically ∫2xex dx\int 2x e^x\,dx). Because the integral is simplified from x2x^2 to xx, it is not yet completely solved, and integration by parts must be applied once more to eliminate xx.

  • Question: What would happen if u=exu = e^x were chosen instead of u=x2u = x^2 for that problem?

  • Response: Choosing u=exu = e^x would make the integral significantly worse. The power of xx in dvdv would integrate up to x33\frac{x^3}{3}, increasing the overall complexity of the integrand rather than reducing it.