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Code Scholars Tutoring · AP Calculus BC

Integration by parts: choosing what to differentiate

Unit 6 · Topics 6.11 · Detailed notes, a worked example, and a self-check with reasoning.

Understand the idea

Integration by parts reverses the product rule. It is useful when an integrand contains a product whose factors play different roles: differentiating one factor simplifies it, while the other has an antiderivative you can find. The goal is to replace the original integral with an easier one, not simply to make it look different.

Choose u\displaystyle u and dv\displaystyle dv together. A polynomial often becomes simpler after differentiation, while exponential and trigonometric factors are usually manageable to integrate. Logarithms can be good choices for u\displaystyle u even when the original integrand does not visibly look like a product: write ln⁡x=(ln⁡x)⋅1\displaystyle \ln x=(\ln x)\cdot1.

For definite integrals, evaluate the product at both bounds and keep the remaining integral on the same interval. For an indefinite integral, add one arbitrary constant at the end. If the remaining integral is harder, reconsider the choice of u\displaystyle u before continuing.

∫u dv=uv−∫v du,∫abu dv=[uv]ab−∫abv du\int u\,dv=uv-\int v\,du,\qquad \int_a^b u\,dv=[uv]_a^b-\int_a^b v\,du

A method you can apply

  1. Identify a product and choose u\displaystyle u so that du\displaystyle du simplifies it.
  2. Integrate dv\displaystyle dv carefully to obtain v\displaystyle v, including any scale factor.
  3. Substitute into uv−∫v du\displaystyle uv-\int v\,du, finish the remaining integral, and differentiate to check.

Worked example

Find ∫xe2x dx\displaystyle \int x e^{2x}\,dx.

  1. Take u=x\displaystyle u=x and dv=e2x dx\displaystyle dv=e^{2x}\,dx. Then du=dx\displaystyle du=dx and v=12e2x\displaystyle v=\tfrac12e^{2x}.
  2. Integration by parts gives 12xe2x−12∫e2x dx\displaystyle \tfrac12xe^{2x}-\tfrac12\int e^{2x}\,dx.
  3. The remaining integral contributes 14e2x\displaystyle \tfrac14e^{2x}, which must be subtracted.
  4. Differentiating the resulting expression produces xe2x\displaystyle xe^{2x}; the extra exponential terms cancel.
The antiderivative is 12xe2x−14e2x+C\displaystyle \tfrac12xe^{2x}-\tfrac14e^{2x}+C.

Try it yourself

Find ∫ln⁡x dx\displaystyle \int\ln x\,dx for x>0\displaystyle x>0.

Show solution and reasoning

Set u=ln⁡x\displaystyle u=\ln x and dv=dx\displaystyle dv=dx. Then du=dx/x\displaystyle du=dx/x and v=x\displaystyle v=x, so the result is xln⁡x−∫1 dx=xln⁡x−x+C\displaystyle x\ln x-\int1\,dx=x\ln x-x+C. Its derivative is ln⁡x\displaystyle \ln x.

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