Question Statement

Evaluate the integral:

$\int \frac{1}{x^2-9}dx$

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Background and Explanation

This problem requires integrating a rational function where the denominator can be factored as a difference of squares. The method of partial fraction decomposition is used to break the complex fraction into simpler terms that can be integrated individually using basic logarithmic rules.

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Solution

We begin by recognizing that the denominator $x^2-9$ is a difference of squares, which factors into $(x-3)(x+3)$. This allows us to use partial fraction decomposition.

Let:

$I=\int \frac{1}{x^2-9}dx=\int \frac{1}{(x{)}^2-(3{)}^2}dx=\int \frac{1}{(x-3)(x+3)}dx$

To decompose the integrand, we set up partial fractions:

$\frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}$

Multiplying both sides by $(x-3)(x+3)$ to clear the denominators:

$1=A(x+3)+B(x-3)$

Finding constant $A$: Put $x-3=0$, which means $x=3$, in equation (2):

$\begin{array}{rl}1& =A(3+3)+B(3-3)\\ 1& =A(6)+B(0)\\ 1& =6A\\ A& =\frac{1}{6}\end{array}$

Finding constant $B$: Put $x+3=0$, which means $x=-3$, in equation (2):

$\begin{array}{rl}1& =A(-3+3)+B(-3-3)\\ 1& =A(0)+B(-6)\\ 1& =-6B\\ B& =-\frac{1}{6}\end{array}$

Substituting back: Putting the values of $A$ and $B$ into equation (1):

$\frac{1}{(x-3)(x+3)}=\frac{\frac{1}{6}}{x-3}+\frac{-\frac{1}{6}}{x+3}=\frac{1}{6(x-3)}-\frac{1}{6(x+3)}$

Integrating: Now we integrate both sides with respect to $x$:

$\begin{array}{rl}\int \frac{1}{(x-3)(x+3)}dx& =\frac{1}{6}\int \frac{1}{x-3}dx-\frac{1}{6}\int \frac{1}{x+3}dx\\ \int \frac{1}{x^2-9}dx& =\frac{1}{6}\ln |x-3|-\frac{1}{6}\ln |x+3|+C\\ & =\frac{1}{6}\left[\ln |x-3|-\ln |x+3|\right]+C\\ & =\frac{1}{6}\ln \left|\frac{x-3}{x+3}\right|+C\end{array}$

(Note: Absolute value bars are included since the logarithm is only defined for positive arguments, and the domain of the original integrand excludes $x=\pm 3$.)

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Key Formulas or Methods Used

• Difference of Squares: $a^2-b^2=(a-b)(a+b)$
• Partial Fraction Decomposition: $\frac{1}{(x-a)(x+b)}=\frac{A}{x-a}+\frac{B}{x+b}$
• Logarithmic Integration: $\int \frac{1}{x}dx=\ln |x|+C$
• Logarithm Properties: $\ln a-\ln b=\ln \left(\frac{a}{b}\right)$

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Summary of Steps

1. Factor the denominator: Recognize $x^2-9$ as $(x-3)(x+3)$
2. Set up partial fractions: Write $\frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}$
3. Clear denominators: Multiply through by $(x-3)(x+3)$ to get $1=A(x+3)+B(x-3)$
4. Solve for constants:
   • Substitute $x=3$ to find $A=\frac{1}{6}$
   • Substitute $x=-3$ to find $B=-\frac{1}{6}$
5. Rewrite integrand: Express as $\frac{1}{6(x-3)}-\frac{1}{6(x+3)}$
6. Integrate term by term: Apply $\int \frac{1}{u}du=\ln |u|$ to each fraction
7. Combine results: Factor out $\frac{1}{6}$ and use logarithm properties to simplify to $\frac{1}{6}\ln \left|\frac{x-3}{x+3}\right|+C$