Anand Sir · Learn with understanding

Integration by parts: choosing u and checking your answer

Work through integration by parts with x eˣ and ln x, learn how to choose u, check by differentiation and try a practice question with a worked solution.

Integration by parts reverses the product rule. It is useful when differentiating one factor makes a product simpler and the other factor is easy to integrate. The choice of factors matters: the aim is a simpler remaining integral.

These are original teaching examples for students practising calculus, including IB Maths AA HL. They are not past-paper questions. You should already be comfortable with the product rule, basic integration and the derivative of eˣ.

Where does the formula come from?

The product rule says (uv)′ = u′v + uv′. Integrate both sides and rearrange:

∫ u dv = uv − ∫ v du.

Choose u to differentiate and dv to integrate. Write du and v explicitly before substitution; this helps prevent sign and factor errors.

Worked example: integrate x eˣ

Choose u = x and dv = eˣ dx. Then du = dx and v = eˣ. Differentiating x removes the polynomial factor, making the remaining integral easier:

∫ x eˣ dx = x eˣ − ∫ eˣ dx
= x eˣ − eˣ + C
= (x − 1)eˣ + C.

Check by differentiating: the derivative of (x − 1)eˣ is eˣ + (x − 1)eˣ = x eˣ, which recovers the original integrand.

Why not choose u = eˣ?

Taking dv = x dx gives v = x²/2. The remaining integral then contains x²eˣ, which is harder than the original xeˣ. A valid application of the formula is not always a useful choice.

With limits from 0 to 1

Use the antiderivative, then subtract the lower-bound value from the upper-bound value:

∫₀¹ x eˣ dx = [(x − 1)eˣ]₀¹ = 0 − (−1) = 1.

There is no + C in the final numerical answer. The integrand is non-negative on this interval, so a negative result would signal an error.

Worked example: integrate ln x, for x > 0

Write the integrand as 1 × ln x. Choose u = ln x and dv = dx, giving du = (1/x) dx and v = x.

∫ ln x dx = x ln x − ∫ x(1/x) dx
= x ln x − x + C.

The derivative of x ln x − x is ln x + 1 − 1 = ln x. The domain x > 0 matters because this example uses the real logarithm ln x.

Common errors to catch

  • Losing the minus sign: the formula subtracts the remaining integral.
  • Confusing dv with v: integrate dv first; do not copy the integrand unchanged unless its antiderivative really is the same.
  • Forgetting a chain-rule factor: ∫ e²ˣ dx is e²ˣ/2, not e²ˣ.
  • Stopping after one application: a higher-degree polynomial may require integration by parts more than once.

Try it yourself: integrate x e²ˣ

Choose u and dv, calculate du and v, then check your answer by differentiating. Attempt this before opening the solution.

Show the worked solution

Set u = x and dv = e²ˣ dx. Then du = dx and v = e²ˣ/2.

∫ x e²ˣ dx = x e²ˣ/2 − (1/2)∫ e²ˣ dx
= x e²ˣ/2 − e²ˣ/4 + C.

Differentiation gives e²ˣ/2 + x e²ˣ − e²ˣ/2 = x e²ˣ. The two extra terms cancel.

Choose your next step

Explain why your choice of u simplified the question. For another reasoning task, try our proof by induction guide. For course-specific support, explore IB Maths AA HL tutoring and lesson pricing, then bring your attempted working when you enquire.

Check the official IB Mathematics course information and your school's guide for your examination cohort. This page teaches the method rather than reproducing an examination specification.