Integration of Rational Functions

In this section we will take a more detailed look at the use of partial fraction decomposi-tions in evaluating integrals of rational functions, a technique we first encountered in the inhibited growth model example in the previous section. However, we will not be able to complete the story until after the introduction of the inverse tangent function in Section 6.5. We begin with a few examples to illustrate how some integration problems involving rational functions may be simplified either by a long division or by a simple substitution. Example To evaluate x 2 x + 1 dx, we first perform a long division of x + 1 into x 2 to obtain x 2 x + 1 = x − 1 + 1 x + 1. Then x 2 x + 1 dx = x − 1 + 1 x + 1 dx = 1 2 x 2 − x + log |x + 1| + c. Example To evaluate 2x + 1 x 2 + x dx, we make the substitution u = x 2 + x du = (2x + 1)dx. Then 2x + 1 x 2 + x dx = 1 u du = log |u| + c = log |x 2 + x| + c. Example To evaluate x x + 1 dx, we perform a long division of x + 1 into x to obtain x x + 1 = 1 − 1 x + 1. Then x x + 1 dx = 1 − 1 x + 1 dx = x − log |x + 1| + c. Alternatively, we could evaluate this integral with the substitution u = x + 1 du = dx.