Q.Integrate the following function: $\frac{1}{x^2 - 9}$

Concept: Partial Fraction Decomposition.

First, factor the denominator: $x^2 - 9 = (x-3)(x+3)$.

We set up the decomposition:

Multiply through by $(x-3)(x+3)$:

Solve for $A$ and $B$.

Let $x = 3$: $1 = A(6) \Rightarrow A = \frac{1}{6}$.

Let $x = -3$: $1 = B(-6) \Rightarrow B = -\frac{1}{6}$.

Thus:

Now integrate term by term:

The key here is that $x^2 - 9$ factors as $(x-3)(x+3)$, a product of two distinct linear factors. When you have a rational function like this — a constant numerator over a factorable quadratic denominator — partial fraction decomposition is the natural tool. The idea is to break the complicated fraction into a sum of simpler fractions, each with a single linear denominator, which we can integrate directly using the natural logarithm.

Why does this work? Because integration is linear: the integral of a sum is the sum of the integrals. And each piece $\frac{A}{x-a}$ integrates to $A\log|x-a|$. So if we can find the right constants $A$ and $B$, the problem reduces to two easy log integrals.

Let’s do it step by step.

1. Factor the denominator and set up the decomposition. Since $x^2 - 9 = (x-3)(x+3)$, we write:

where $A$ and $B$ are constants we need to find.

2. Clear the denominators. Multiply both sides by $(x-3)(x+3)$:

This equation must hold for all $x$ (except $x = \pm 3$, where the original fraction is undefined, but the identity holds algebraically).

3. Solve for $A$ and $B$.

   There are two efficient methods. I’ll show both — pick whichever you find clearer.

   Method 1: Substitution (the “cover-up” trick).

   Choose $x = 3$ to make the $B$ term vanish:

Choose $x = -3$ to make the $A$ term vanish:

Method 2: Equating coefficients.

Expand the right side: $1 = A x + 3A + B x - 3B = (A+B)x + (3A - 3B)$.

Compare coefficients of $x$ and the constant term:

From the first equation, $B = -A$. Substitute into the second: $3A - 3(-A) = 6A = 1$, so $A = \frac{1}{6}$, and then $B = -\frac{1}{6}$. Same result.

4. Write the decomposed form. Now we have:

5. Integrate term by term.

Each integral is a standard log form: $\int \frac{1}{u} du = \log|u| + C$. So:

6. Simplify using logarithm properties. Combine the two logs into one:

Method: Factor first, then partial fractions ($\dfrac{1}{x^2-a^2}$ type)

Use this whenever the denominator is written as an unfactored quadratic like $x^2-a^2$. Factor it into distinct linear pieces, then decompose into logarithms.

Steps

Step 1: Factor the denominator. A difference of squares splits as $x^2-a^2=(x-a)(x+a)$.

Step 2: Set up one constant per linear factor.

Step 3: Solve by cover-up — put $x=a$ and $x=-a$ to isolate $A$ and $B$.

Step 4: Integrate and combine.

which for this symmetric case collapses to $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+C$.

Common Mistakes

Mistake 1: Trying to integrate $\frac{1}{x^2-9}$ before factoring.

Why it's wrong: $\frac{1}{x^2-9}$ is not a standard form on its own; only after writing it as $\frac{1}{(x-3)(x+3)}$ can partial fractions apply. Correct approach: factor the difference of squares first.

Mistake 2: Confusing it with the arctangent form.

Why it's wrong: $\frac{1}{x^2-9}$ has real roots and gives logarithms, whereas $\frac{1}{x^2+9}$ (no real roots) gives $\frac13\tan^{-1}\frac x3$. Correct approach: check the sign — a minus makes it a log, a plus makes it an arctan.

Mistake 3: Getting the $\tfrac{1}{6}$ factor wrong.

Why it's wrong: $A=\frac16,\ B=-\frac16$; forgetting the $\frac{1}{2a}=\frac16$ scaling gives an answer off by a constant multiple. Correct approach: solve $1=A(x+3)+B(x-3)$ at the roots to fix the constants exactly.