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Integration methods

Integration by Parts: Choosing u and dv

Integration by parts reverses the product rule. It is useful when one factor simplifies under differentiation and the other is easy to integrate.

Direct answer

∫u dv\int u\,\mathrm{d}v

Choose dv so it can be integrated and u so differentiation makes the remaining integral simpler.

When to use integration by parts

Choose the method from the top-level mathematical structure before manipulating symbols. Conditions are part of the task, not optional notes.

Ask first 1

Which structure controls the method?

What to do

Use parts for a product with a simplifying derivative factor and an integrable companion factor.

Ask first 2

Which condition can change the answer?

What to do

Choose dv so it can be integrated and u so differentiation makes the remaining integral simpler.

Ask first 3

How should the result be checked?

What to do

Differentiate the completed expression and confirm product terms cancel correctly.

Rearrange the product rule

Start from d(uv)=u dv+v du.

Integrate both sides over the same interval or as antiderivatives.

Move ∫v du to obtain ∫u dv=uv−∫v du.

If the new integral is harder, reconsider the choice or try substitution.

1
d(uv)=u dv+v du
2
uv=∫u dv+∫v du
3
∫u dv\int u\,\mathrm{d}v

Worked examples

∫x exp(x) dx
ex (x−1)+Ce^{x}\,\left(x - 1\right) + C

Choose u=x and dv=exp(x)dx.

∫x cos(x) dx
x sin(x)+cos(x)+C

Choose u=x and dv=cos(x)dx.

∫ln⁡x dx\int \ln x\,\mathrm{d}x
x ln(x)−x+C

Treat ln(x) as ln(x)·1 for x>0.

Common mistakes

  • • Losing the subtraction sign.
  • • Choosing a hard dv.
  • • Stopping before evaluating the remaining integral.
  • • Using parts when substitution is simpler.

Error clinic: locate the first invalid step

Incorrect attempt

∫x exp(x)dx=x exp(x)+exp(x)

Why it fails

The remaining term must be subtracted.

Correction

x exp(x)−exp(x)+C

Incorrect attempt

∫ln(x)dx=x ln(x)

Why it fails

Differentiation gives ln(x)+1.

Correction

x ln(x)−x+C

Practice the same idea with a changed structure

Name the rule and its conditions before revealing the reference answer. Explain every sign and factor.

∫x cos(x) dx

Check before solving

Choose u=x and dv=cos(x)dx.

Reveal the reference answer

Reference answer

x sin(x)+cos(x)+C
∫ln⁡x dx\int \ln x\,\mathrm{d}x

Check before solving

Treat ln(x) as ln(x)·1 for x>0.

Reveal the reference answer

Reference answer

x ln(x)−x+C

Try the worked examples in the calculator

Each link opens the matching calculator with the expression, variable and required conditions already filled in.

Integrais:x exx\,e^{x}Integrais:x cos⁡xx\,\cos x
Open the calculator without a preset: Integrais

Frequently asked questions

What should be u?+

Prefer a factor that becomes simpler when differentiated.

Can parts be repeated?+

Yes. Polynomial times exponential or trigonometric expressions often need repeated steps.

How do I verify the sign?+

Differentiate uv−∫vdu and confirm the integrand returns.

Conclusion

Choose u and dv to simplify the remaining integral, retain the minus sign, and verify by differentiation.

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