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:

\int\frac{(x-1)\,dx}{x^3-x^2-2x}

Solution

(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:

I=\int\frac{(x-1)\,dx}{x(x^2-x-2)}=\int\frac{(x-1)\,dx}{x(x-2)(x+1)}

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:

Q(x)=(a_1x+b_1)(a_2x+b_2)\cdots(a_nx+b_n)

(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})

Then

\frac{P(x)}{Q(x)}=\frac{A_1}{a_1x+b_1}+\frac{A_2}{a_2x+b_2}+\cdots+\frac{A_n}{a_nx+b_n}

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


Now

\frac{(x-1)}{x(x-2)(x+1)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+1}

=\frac{A(x-2)(x+1)+Bx(x+1)+Cx(x-2)}{x(x-2)(x+1)}

\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:

\underline{x=0},

0-1=A(-2)(1)+B(0)(1)+C(0)(-2)

\Rightarrow A=1/2

\underline{x=-1},

-1-1=A(-3)(0)+B(-1)(0)+C(-1)(-3)

\Rightarrow C=-2/3

\underline{x=2},

2-1=A(0)(3)+B(2)(3)+C(2)(0)

\Rightarrow B=1/6

 

\frac{(x-1)}{x(x-2)(x+1)}=\frac{1}{2}.\frac{1}{x}+\frac{1}{6}.\frac{1}{x-2}+\frac{-2}{3}\frac{C}{x+1}

\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

=\frac{1}{2}\int\frac{1}{x}\,dx+\frac{1}{6}\int\frac{1}{x-2}\,dx+\frac{-2}{3}\int\frac{1}{x+1}\,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:

=\frac{1}{2}\int\frac{1}{x}\,dx+\frac{1}{6}\int\frac{1}{u}\,du+\frac{-2}{3}\int\frac{1}{v}\,dv

=\frac{1}{2}\ln|x|+\frac{1}{6}\ln|u|+\frac{-2}{3}\ln|v|+C

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

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

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

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

I=\frac{1}{6}\left(\ln|x^3|+\ln|x-2|+\ln\left(\frac{1}{|x+1|^4}\right)\right)+C

=\frac{1}{6}\ln\left(|x^3||x-2|\frac{1}{|x+1|^4}\right)+C

=\frac{1}{6}\ln\left|\frac{x^3(x-2)}{(x+1)^4}\right|+C

\Box

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

f(x)=\frac{1}{(x-a)(x-b)}

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

f(x)=\frac{1}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}

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}:

1=\frac{1}{a-b}(x-b)+\frac{1}{b-a}(x-a)

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:

\frac{1}{(x-a)(x-b)}=\frac{1/(a-b)}{x-a}+\frac{1/(b-a)}{x-b}

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

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