Q.Evaluate the definite integral: $\int_2^3 \frac{dx}{x^2-1}$

The key idea is Partial Fraction Decomposition to split the rational function into simpler terms.

First, factor the denominator: $x^2 - 1 = (x-1)(x+1)$.

Write $\frac{1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1}$. Solving gives $A = \frac12$, $B = -\frac12$.

So the integral becomes:

$\int_2^3 \frac{dx}{x^2-1} = \frac12 \int_2^3 \left( \frac{1}{x-1} - \frac{1}{x+1} \right) dx$ …

The integrand $\frac{1}{x^2-1}$ is a rational function where the denominator factors as $(x-1)(x+1)$. Direct integration isn't possible, but we can rewrite it as a sum of two simpler fractions — each of the form $\frac{A}{x-1} + \frac{B}{x+1}$ — which integrate directly to logarithms. This is the essence of partial fraction decomposition.

Let’s work through it.

1. Factor the denominator and set up the decomposition $x^2 - 1 = (x-1)(x+1)$. We assume constants $A$ and $B$ such that:

2. Clear denominators Multiply both sides by $(x-1)(x+1)$:

3. Solve for $A$ and $B$ Expand: $1 = Ax + A + Bx - B = (A+B)x + (A-B)$. For this to hold for all $x$, coefficients of $x$ and the constant term must match:

Adding the equations gives $2A = 1 \Rightarrow A = \frac{1}{2}$. Then $B = -\frac{1}{2}$.

4. Rewrite the integral

5. Integrate each term $\int \frac{dx}{x-a} = \log|x-a| + C$, so:

6. Evaluate the limits …

Method: Definite integral by partial fractions

When the integrand is a proper rational function with a factorable denominator, e.g. $\dfrac{1}{x^{2}-1}$, split into partial fractions, integrate to logarithms, then evaluate.

Steps

Step 1: Factor the denominator. $x^{2}-1=(x-1)(x+1)$.

Step 2: Decompose.

Step 3: Integrate each piece. …

Common Mistakes

Mistake 1: Forgetting the $\dfrac12$ from the partial-fraction coefficients.

Why it's wrong: $\dfrac{1}{(x-1)(x+1)}=\dfrac12\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)$; dropping the $\tfrac12$ doubles the answer. Correct approach: solve for the constants ($A=\tfrac12,\,B=-\tfrac12$).

Mistake 2: Integrating $\dfrac{1}{x^{2}-1}$ as $\tan^{-1}$ or $\dfrac{1}{x^{2}-1}\to\log(x^{2}-1)$.

Why it's wrong: $x^{2}-1$ (a difference) gives a log-of-ratio via partial fractions, not an arctangent, and not $\log(x^{2}-1)$. Correct approach: decompose first. …