Integration by Parts

There's a magic Integration by Parts formula:
[A]     ∫ u dv = u v - ∫ v du

The format is meant (I assume) for easy memorization.

After having taught this stuff for some 30 years, the "best" way (for me) to teach integration by parts goes something like this:

  • Rewrite the above ritual like so:
    ∫ u (dv/dx) dx = u v - ∫ v (du/dx).
  • Recognize that you've replaced the integration of some product (that's u (dv/dx)) by some other product (that's v (du/dx)).
    That is, the one factor (u) gets differentiated and the other factor (dv/dx) gets integrated.
  • You stare at an integration problem, split it up into two factors and ask yourself:
    Which looks nicer?
    1. Integrating (first factor differentiated) x (second factor integrated)
      or
    2. Integrating (second factor differentiated) x (first factor integrated)
  • Having decided upon which you'd like to reappear differentiated, let u be the factor you want to integrate!
    Everthing else under the integral sign is then dv ... then use [A].

Examples:
You're given: ∫ x2ex dx.
Differentiatng the x2 gets us: ∫ 2x ex dx which looks much nicer than ∫ (1/3) x3 ex dx
Then we let u = x2 and replace the original integral by u v - ∫ v du
Everthing else is dv, so dv = ex dx and we can choose v = ex.

The new problem?
Evaluate:         x2 ex - ∫ 2x ex dx
(Here, you'll want to do the Integration by Parts routine again.)

You're given: ∫ x2 log(x) dx.
Differentiatng the log(x) gets us: ∫ (1/3) x3 (1/x) dx = ∫ (1/3) x2 dx which looks good.
(The other option looks unsanitary!!)
Then we let u = log(x) and replace the original integral by u v - ∫ v du
Note that dv = x2 dx so we can choose v = x3/3.

The new problem?
Evaluate:     log(x) x3/3 - ∫ (1/3) x2 dx
(Piece o' cake.)

You're given: : ∫ x3 cos(x) dx.
Differentiating the x3 gets us: ∫ 3 x2 sin(x) dx which is a step in the right direction.
Then we let u = x3 and replace the original integral by u v - ∫ v du
Note that dv = cos(x) dx so we can choose v = sin(x).

The new problem?
Evaluate:         x3 sin(x) - ∫ 3 x2 sin(x) dx
(At least the exponent is smaller ... so do it again.)

You're given: ∫ arctan(x) dx where we recognize the 2 factors: arctan(x) and 1.
(You're now tempted to differentiate the "1", right? Try letting u = 1.)
Differentiating arctan(x) gets us: ∫ 1/(1+x2) x dx (where we integrated "1" to give "x").
Then we let u = arctan(x) and replace the original integral by u v - ∫ v du
Note that dv = dx so we can choose v = x.

The new problem?
Evaluate:         x arctan(x) - ∫ x / (1 + x2) dx