Math Help

Integrals Involving Rational Functions

Example 1: Partial Fractions with Distinct Linear Factors

Note: P(x) = 1 and Q(x) = x² + x - 2 So long division is unnecessary.

I = ∫ dx / (x² + x - 2)

1/ (x² + x - 2) = 1/(x – 1)(x + 2) = A/(x – 1) + B/(x + 2)

The terms A/(x – 1) and B/(x + 2) are the “partial fractions”

Next put them under a common denominator. The result is:

1/ (x² + x - 2) = [A(x + 2) + B(x – 1)] / (x – 1)(x + 2)

and equate numerators.

1 = A(x + 2) + B(x – 1) = x(A + B) + (2A – B)

Solve for A and B. There are two methods to obtain the solution.

Method 1: Equate terms of equal powers of x on both sides of the equal sign.

The result of this yields two equations in the unknowns A and B as follows:

2A – B = 1 and A + B = 0 Solve for A and B. A = 1/3, B = -1/3

Method 2: Both sides of the equation 1 = A(x + 2) + B(x – 1)

must hold for any value of x. So pick values of x that enable you to solve

most easily for A and B.

For example pick x = 1, which gets rid of B. Then you get 1 = 3A. so A = 1/3.

Next pick x = -2 which gets rid of A. Then 1 = -3B and B = - 1/3.

Note: This method may be quicker and easier.

So I = (1/3) ∫ dx / (x – 1) - (1/3) ∫ dx /(x + 2)

and I = (1/3) ln| x – 1| - (1/3) ln|x + 2| + C (result)

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