Chapter 6.5, Problem 8E

To determine

To find: The value of the integral.

Answer to Problem 8E

The integral is:

  $x+\frac{\ln \frac{|x-3|}{|x+3|}}{2}+C$

Explanation of Solution

Given information:

The integral is: $\int \frac{x^2-6}{x^2-9}dx$

Calculation:

The given integral is: $\int \frac{x^2-6}{x^2-9}dx$

  $\frac{x^2-6}{x^2-9}=1+\frac{3}{x^2-9}$

The second fraction can now be divided into its suitable partial fraction configuration.

  $\begin{array}{c}\int \left(1+\frac{3}{x^2-9}\right)dx\\ =\int 1dx+\int \frac{3}{x^2-9}dx\\ =\int 1dx+\int \frac{A^{\prime }}{x+3}dx+\int \frac{B}{x-3}dx\end{array}$

Solve A and B:

  $\begin{array}{c}A(x-3)+B(x+3)=3\\ Ax-3A+Bx+3B=3\end{array}$

Setup a equations:

  $A+B=0$ (Equation 1)

  $-3A+3B=3$ (Equation 2)

Substitution:

(Equation 1) $A=0-B$

(Equation 2):

  $\begin{array}{c}-3(-B)+3B=3\\ \to 3B+3B=3\\ \to 6B=3\\ \to B=\frac{1}{2}\end{array}$

Put the value of B into the Equation 1:

  $\begin{array}{c}A+\left(\frac{1}{2}\right)=0\\ A=-\frac{1}{2}\end{array}$

Reconnect A and B to the integral and get the solution.

  $\begin{array}{c}\int 1dx-\int \frac{1}{2(x+3)}dx+\int \frac{1}{2(x-3)}dx\\ =\int 1dx-\frac{1}{2}\int \frac{1}{x+3}dx+\frac{1}{2}\int \frac{1}{x-3}dx\\ =x-\frac{1}{2}\ln |x+3|+\frac{1}{2}\ln |x-3|+C\\ =x+\frac{\ln |x-3|-\ln |x+3|}{2}+C\\ =x+\frac{\ln \frac{|x-3|}{|x+3|}}{2}+C\end{array}$

Therefore the required value of the given integral $\int \frac{x^2-6}{x^2-9}dx$ is

  $x+\frac{\ln \frac{|x-3|}{|x+3|}}{2}+C$