The integrand is an improper rational function. By "long division" of polynomials, we can rewrite the integrand as the sum of a polynomial and a proper rational function "remainder":

3x + 1                      x2-x-6 ø 3x3 - 2x2 - 19x - 7 3x3 - 3x2 - 18x x2 - x - 7 x2 - x - 6 - 1

So

This looks much easier to work with! We can integrate 3x+1 immediately, but what about -1/(x²-x-6)?

Notice that

This is called the Partial Fraction Decomposition for -1/(x²-x-6).

Our goal now is to determine A and B. Multiplying both sides of the equation by (x+2)(x-3) to clear the fractions,

There are two methods for solving for A and B:

Method 1		Method 2
Collect like terms on the right:		The equation holds for all x.
-1 = (A+B)x+(-3A+2B).		Let x = -2:
Now equate coefficients of		-1 = A(-2-3)+B(-2+2)
corresponding powers of x:		-1 = -5A      ® A = 1/5.
A+B = 0,    -3A+2B = -1.		Now let x = 3:
Solving this system,		-1 = A(3-3)+B(3+2)
A = 1/5, B = -1/5.		-1 = 5B   ® B = -1/5.

So

Returning to the original integral,

In the next example, we have repeated factors in the denominator, as well as an irreducible quadratic factor.

Example

We will evaluate

The integrand is a proper rational function, which we would like to decompose into proper rational functions of the form

[Notice that we have two factors of x in the denominator of the integrand, leading to terms of the form ^A/_x and B/(x²) in the decomposition. The factor x²+x+1 is irreducible and quadratic, so any proper rational function with x²+x+1 as denominator has the form (Cx+D)/( x²+x+1) where C or D may be 0.]

Set

Multiplying through by x²(x²+x+1),

Since x²+x+1 has no real roots, it is easiest to solve for A and B using Method 1:

Collecting like terms on the right,

Equating corresponding powers of x,

So