Indefinite Integrals - Choosing an Integration Method

Choosing an Integration Method

We've been learning the different methods of integration in a very artificial environment. We know that if we're in the "Integration by Substitution" section, we use substitution. If we're in the "Integration by Parts" section, we use integration by parts.

In the real world (by which we mean on exams), the directions probably won't say which method to use—we'll have to figure that ourselves.

The more we practice, the better we'll get at figuring out which method to use. We won't even have to think about it. In the meantime, we have some hints.

• Integration by Substitution

  Use substitution when the integrand can be factored into something with an "inside function" u and something that's more-or-less the derivative of u (if the constant coefficients don't quite agree, that's ok).

  Sample Problem

  We would use integration by substitution on

  

  because x is a constant multiple of the derivative of x², which is an inside function:

  

  Sample Problem

  We can't use substitution on

  

  If we try to let u = x² it just doesn't work, because we have an extra factor of x hanging around:

  

• Integration by Parts

  Use integration by parts when the integrand factors into two things that both include the variable, but integration by substitution doesn't work!

  Sample Problem

  We can use integration by parts on

  

  because we can factor the x² to get

  

  and choose u = x and v' = xe^x2.

  Sample Problem

  We wouldn't use integration by parts on

  

  because this integral begs for substitution.

• Integration by Partial Fractions

  Use the partial fractions technique when you're asked to evaluate a rational function that
  

  • has a lower degree in the numerator than in the denominator, and
      
  • has a denominator that can be factored into distinct linear factors.

  Sample Problem

  We can use the method of partial fractions on

  

  because the numerator has degree 0, the denominator has degree 2, and the denominator factors into

  x² – 2x – 3 = (x – 3)(x + 1).

  Sample Problem

  We wouldn't use the method of partial fractions on

  

  because the denominator factors into

  x² + 2x + 1 = (x + 1)(x + 1).

  These are not distinct linear factors.

  Actually, it is possible to use the method of partial fractions on this example, but the setup is a bit more complicated. We'll stick to the simpler examples of integration by partial fractions.

• Thinking Backwards

  Don't forget the first method we learned to find integrals: "thinking backwards." Sometimes you don't need substitution, parts, or partial fractions—you can simplify the integral and immediately see what to do with it.

  Sample Problem

  We don't need anything fancy to find

  

  Simplify the integral by squaring the integrand and then separating it out:

  

  Then integrate each term:

  

  Sample Problem

  Depending on how comfortable you are with thinking backwards, you might be able to do this one in your head:

  

  However, you're still doing substitution behind the scenes, letting u = 2x + 3.

  Sample Problem

  There's no reasonable way to think backwards from

  

  That's what we learned integration by parts for.