Partial Fractions - Integration techniques

In this part we are going to take a look at integrals of rational expressions of polynomials and again let's start this section out with an integral which we can already do so we can contrast it with the integrals that we'll be doing in this segment.

 ∫ (2x-1 / x2 -x - 6) (dx)

∫ (1/u) (du)

By using u = x² - x - 6 and du = (2x-1) dx

= 1n |x² - x - 6 | + c

Thus, if the numerator is the derivative of the denominator (or a constant multiple of the derivative of the denominator) doing this type of integral is fairly simple.  Though, frequently the numerator isn't the derivative of the denominator (or a constant multiple).  For instance, consider the following integral.

∫ (3x+11/(x²-x-6)) (dx)

In this type of case the numerator is certainly not the derivative of the denominator nor is it a constant multiple of the derivative of the denominator. Hence, the simple substitution which we used above won't work.  Though, if we notice that the integrand can be broken up as follows,

3x + 11 /x²-x-6

= 4/x-3 - 1/x+2

Then the integral is in fact quite simple.