Let’s work step by step to solve this improper integral problem.

Part 1:

We are tasked to evaluate the definite integral
$A(t) = \int_{4}^{t} \frac{1}{x^2 - 2x - 3} dx$ for $t > 4$.

Step 1: Factorize the denominator

The denominator $x^2 - 2x - 3$ can be factorized as:
$x^2 - 2x - 3 = (x - 3)(x + 1)$

Step 2: Use partial fraction decomposition

We write: $\frac{1}{x^2 - 2x - 3} = \frac{A}{x - 3} + \frac{B}{x + 1}$ Multiplying through by $(x - 3)(x + 1)$, we get: $1 = A(x + 1) + B(x - 3)$ Expanding and matching coefficients: $1 = A x + A + B x - 3B \implies (A + B)x + (A - 3B) = 1$ From this:

1. $A + B = 0$
2. $A - 3B = 1$

Solving these equations:

1. $A = -B$
2. Substituting $A = -B$ into $A - 3B = 1$:
   $-B - 3B = 1 \implies -4B = 1 \implies B = -\frac{1}{4}$
   Therefore, $A = \frac{1}{4}$.

Thus: $\frac{1}{x^2 - 2x - 3} = \frac{1/4}{x - 3} - \frac{1/4}{x + 1}$

Step 3: Integrate

$A(t) = \int_{4}^{t} \left( \frac{1/4}{x - 3} - \frac{1/4}{x + 1} \right) dx$ Separate the integrals: $A(t) = \frac{1}{4} \int_{4}^{t} \frac{1}{x - 3} dx - \frac{1}{4} \int_{4}^{t} \frac{1}{x + 1} dx$ The integrals are: $\int \frac{1}{x - 3} dx = \ln|x - 3|, \quad \int \frac{1}{x + 1} dx = \ln|x + 1|$ So: [ A(t) = \frac{1}{4} \left[ \ln|x - 3| \right]{4}^{t} - \frac{1}{4} \left[ \ln|x + 1| \right]{4}^{t} ]

Evaluate: $A(t) = \frac{1}{4} \left( \ln|t - 3| - \ln|4 - 3| \right) - \frac{1}{4} \left( \ln|t + 1| - \ln|4 + 1| \right)$ Simplify: $A(t) = \frac{1}{4} \ln|t - 3| - \frac{1}{4} \ln 1 - \frac{1}{4} \ln|t + 1| + \frac{1}{4} \ln 5$ Since $\ln 1 = 0$, this reduces to: $A(t) = \frac{1}{4} \ln|t - 3| - \frac{1}{4} \ln|t + 1| + \frac{1}{4} \ln 5$ Combine logs: $A(t) = \frac{1}{4} \ln \left( \frac{5(t - 3)}{t + 1} \right)$

Part 2:

Take the limit of $A(t)$ as $t \to \infty$: $\lim_{t \to \infty} A(t) = \lim_{t \to \infty} \frac{1}{4} \ln \left( \frac{5(t - 3)}{t + 1} \right)$ Simplify the fraction inside the log: $\frac{t - 3}{t + 1} = \frac{t(1 - 3/t)}{t(1 + 1/t)} \to \frac{1 - 0}{1 + 0} = 1 \quad \text{as } t \to \infty$ Thus: $\lim_{t \to \infty} A(t) = \frac{1}{4} \ln(5 \cdot 1) = \frac{1}{4} \ln 5$

Part 3:

Since the limit exists and is finite, the improper integral converges, and its value is: $\boxed{\frac{1}{4} \ln 5}$

Would you like me to clarify any step further?

Related Questions:

1. How do you perform partial fraction decomposition in general?
2. What are some strategies to identify if an improper integral diverges?
3. Why is it necessary to evaluate the limit as $t \to \infty$?
4. Can all improper integrals be evaluated using partial fractions?
5. What are other techniques for solving improper integrals (e.g., comparison tests)?

Tip:

When evaluating limits of logarithmic functions, always simplify the argument first to avoid unnecessary complexity!