Learning goals

What you should understand

  • Recognize products for which integration by parts makes the remaining integral simpler.
  • Choose u and dv, then compute du and v without losing the differential.
  • Apply the formula to indefinite and definite integrals and verify the result.
Before you begin

You should be comfortable with the product rule, basic antiderivatives, algebraic simplification, and the difference between definite and indefinite integrals.

The integration-by-parts formula

Starting from d(uv) = u dv + v du and integrating both sides gives the formula. It exchanges one integral for another rather than integrating a product term by term.

∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du
  • Choose u and v with the required derivatives on the interval of interest.
  • For definite integrals, evaluate uv at the same endpoints and preserve the bounds on the remaining integral.

Reusable process

Choose the split that makes progress

  1. 1
    Confirm the product structure

    Look for two factors or rewrite a single logarithm or inverse trigonometric function as one times that function.

  2. 2
    Choose u

    Prefer a factor that becomes simpler when differentiated, such as a logarithm, inverse function, or polynomial.

    u=f(x),du=f′(x) dxu=f(x),\qquad du=f'(x)\,dx
  3. 3
    Choose dv and find v

    The remaining differential must be something you can integrate directly.

    dv=g′(x) dx,v=g(x)dv=g'(x)\,dx,\qquad v=g(x)
  4. 4
    Substitute into the formula

    Write uv first, then subtract the new integral. Check that the new integral is actually simpler.

    uv−∫v duuv-\int v\,du
  5. 5
    Finish and verify

    Integrate the remaining term, simplify, add C when needed, and differentiate the final answer.

Decision check

Which factor should be u?

For the integral below, compare choosing x with choosing e to the x. Which derivative simplifies the product?

∫xex dx\int xe^x\,dx
Choose the factor to use as u

Check the reasoning

Choose u = x and dv = e^x dx. Differentiating x gives 1, while integrating e to the x leaves e to the x. The remaining integral is therefore simpler than the original product.

Worked examples

Build from recognition to verification

Polynomial times exponential

Differentiate the polynomial and integrate the exponential.

∫xex dx\int xe^x\,dx
  1. Choose: Let u = x and dv = e to the x dx.
    u=x, du=dx, dv=ex dx, v=exu=x,\ du=dx,\ dv=e^x\,dx,\ v=e^x
  2. Apply: Substitute into the formula.
    xex−∫ex dxxe^x-\int e^x\,dx
  3. Finish: Integrate the remaining exponential.
    ex(x−1)+Ce^x(x-1)+C

Answer

∫xex dx=ex(x−1)+C\int xe^x\,dx=e^x(x-1)+C

Check: The product rule gives e to the x times x minus 1 plus e to the x, which simplifies to x e to the x.

A definite logarithmic integral

Rewrite ln x as one times ln x, then keep the endpoints through the formula.

∫1eln⁡x dx\int_1^e\ln x\,dx
  1. Choose: Let u be ln x and dv be dx.
    u=ln⁡x, du=1x dx, v=xu=\ln x,\ du=\frac1x\,dx,\ v=x
  2. Apply with bounds: Evaluate the uv term and the remaining integral on the same interval.
    [xln⁡x]1e−∫1e1 dx[x\ln x]_1^e-\int_1^e1\,dx
  3. Evaluate: The first term is e and the remaining integral is e minus 1.
    e−(e−1)=1e-(e-1)=1

Answer

∫1eln⁡x dx=1\int_1^e\ln x\,dx=1

Check: The antiderivative x ln x minus x evaluates to zero at e and minus one at 1, so upper minus lower is one.

Common mistakes

Notice the decision before the algebra

Choosing u by position

Always choosing the first written factor rather than the factor that simplifies when differentiated.

Fix: Test the derivative of each plausible u and the antiderivative of the remaining dv before committing.

Dropping the minus sign

Writing uv plus the remaining integral.

Fix: Derive the formula from the product rule or say uv minus integral v du before substituting.

A harder new integral

Finishing the substitution even though integral v du is more complicated.

Fix: Return to the split and consider substitution, algebraic rewriting, or another choice of u.

Mishandling bounds

Changing to u-bounds for only part of the expression or forgetting to evaluate uv at the endpoints.

Fix: Keep the original variable throughout integration by parts, or change every bounded term consistently.