This question was asked at Monday’s tutorial (10/01/11) but the fire alarm went off mid-solution

Section 6.4, Q. 5

Evaluate the following integral:



(Remarks in italics are by me and would not be required in an exam situation)

Simplify the integrand to get it into a usable form:


Rule 1 (Section 6.4)

Given a rational function P(x)/Q(x) with \text{deg}(P)<\text{deg}(Q), such that Q(x) factors into non-repeated linear terms:


(non-repeated means that no linear term  is equal to a constant multiple of another; e.g. (a_ix+b_i)=k(a_jx+b_j) for i\neq j, k\in \mathbb{C})



for some constants A_1,A_2,\dots,A_n\in\mathbb{C}.




\Rightarrow x-1=A(x-2)(x+1)+Bx(x+1)+Cx(x-2)   (*)

At this point we have two options:

  1. Multiply out and rearrange the right-hand side into a quadratic in x. Compare and set equal the coefficients of x^2, x and the constant terms to generate three simultaneous equations for A,B,C\in\mathbb{C}.
  2. Choose three values of x (say x=0,-1,2 for ease of computation, see the note below) to generate three simultaneous equations for A,B,C\in\mathbb{C}. We’ll use this method here:



\Rightarrow A=1/2



\Rightarrow C=-2/3



\Rightarrow B=1/6



\Rightarrow I=\int\left(\frac{1}{2}.\frac{1}{x}+\frac{1}{6}.\frac{1}{x-2}+\frac{-2}{3}\frac{1}{x+1}\right)\,dx


Making the substitutions u=x-2 and v=x+1 (it’s always a good idea to do this rather than court difficulty with terms of the form 1/(ax+b) with a\neq 1), and noting du=dv=dx:




Recalling \ln a+\ln b=\ln ab, try to write this expression as k\ln |p|+k\ln |q|+k\ln |r| for some p,q,r functions of x:

I =\frac{3}{6}\ln|x|+\frac{1}{6}\ln|x-2|+\frac{-4}{6}\ln|x+1|+C


Now using |x|^n=|x^n|, x^{-n}=1/x^n, and |a||b|=|ab|:





Note: You may notice that in using technique 2. for finding A,B,C that we use values of x for which both sides of (*) are actually undefined. By means of the following general example let me show you why this is not a problem.

Example: Let a,b\in\mathbb{R} such that a\neq b. Find the partial fraction decomposition of


Solution: In essence this question asks us to find real numbers A,B such that:


These expressions must agree at all points except at x=a,b where f(x) is not defined. Suppose we multiply across by (x-a)(x-b):

1=A(x-b)+B(x-a)  (**)

Now look at this equation in isolation, and forgetting about our original problem, we can see that we can find A,B\in\mathbb{R}, such that this equation holds for all x\in\mathbb{R}. It is an easy exercise to show that (**) is solved by: A=1/(a-b) and B=1/(b-a). That is for all x\in\mathbb{R}:


Now if we take this expression, and divide across by (x-a)(x-b) we will get a new expression, which is true as long as we didn’t divide by zero, i.e. (x-a)\neq 0,(x-b)\neq 0\Leftrightarrow x\neq a,b. This statement is:


These functions agree on \mathbb{R}\backslash\{a,b\}, as is our want.