Mathematics II / Integration ① (Indefinite Integrals, Substitution, and Integration by Parts) 3 / 6

Integration by Parts

This formula is used for integrating products. It is the reverse application of the product rule for differentiation.

Formula for Integration by Parts

∫f(x)g'(x)dx = f(x)g(x) − ∫f'(x)g(x)dx

Or ∫uv'dx = uv − ∫u'v dx

Which Function Should Be u?

In general, choose u in the order “polynomial > exponential > trigonometric > logarithmic” (choose the one that becomes simpler when differentiated).

Example ①
Calculate ∫x(x+1)⁴ dx using integration by parts.
Answer:Set u = x and v' = (x+1)⁴. Then v = (x+1)⁵/5。
= x(x+1)⁵/5 − ∫(x+1)⁵/5 dx
= x(x+1)⁵/5 − (x+1)⁶/30 + C = (x+1)⁵(6x−(x+1))/30 + C = (x+1)⁵(5x−1)/30 + C
Example ②
State the method for ∫x × 2ˣ dx (an integral involving an exponential, for reference).
Answer (method):Set u = x (it disappears when differentiated) and v' = 2ˣ (whose integral is 2ˣ/ln2), then use integration by parts.
Within Mathematics II, it is mainly used for integrating polynomials.
Key Points
  • ∫uv' = uv − ∫u'v: choose u so that the integral on the right is simpler than the one on the left
  • The typical form is a polynomial × an integrable function

Practice Problems

  1. Calculate ∫x(x−1)³ dx using integration by parts.
  2. Calculate ∫(x+1)(x+2)² dx using integration by parts (or by expanding directly).
  3. Calculate ∫x² (x+1) dx both by expansion and by integration by parts, and confirm that the results agree.

Answers and Explanations

  1. Answer:u=x、v=(x−1)⁴/4。x(x−1)⁴/4−∫(x−1)⁴/4 dx = x(x−1)⁴/4−(x−1)⁵/20+C
    Explanation: = (x−1)⁴(5x−(x−1))/20+C = (x−1)⁴(4x+1)/20+C。
  2. Answer:Expansion: (x+1)(x²+4x+4)=x³+5x²+8x+4。Integration: x⁴/4+5x³/3+4x²+4x+C
    Explanation: Expansion is faster.
  3. Answer:x⁴/4+x³/3+C (by expansion).
    Explanation: x²(x+1)=x³+x²。∫(x³+x²)dx=x⁴/4+x³/3+C。The same result is obtained by integration by parts.

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