§(x^2 -2x -3/(x -1)(x^2 -4)dx?

To solve this integral, we first need to decompose the rational function into partial fractions.

Given the rational function:

(x^2 - 2x - 3) / ((x - 1)(x^2 - 4))

We need to express it as the sum of two fractions with simpler denominators.

First, factor the denominator:

(x - 1)(x^2 - 4) = (x - 1)(x - 2)(x + 2)

Now, we need to express the numerator as a sum of two fractions with these simpler denominators:

(x^2 - 2x - 3) = A/(x - 1) + B/(x - 2) + C/(x + 2)

Multiplying both sides by the denominator, we get:

x^2 - 2x - 3 = A(x - 2)(x + 2) + B(x - 1)(x + 2) + C(x - 1)(x - 2)

Expanding and equating coefficients, we get:

x^2 - 2x - 3 = A(x^2 - 4) + B(x^2 + x - 2) + C(x^2 - 3x + 2)

Now, we can solve for A, B, and C by comparing coefficients.

After finding the values of A, B, and C, we can rewrite the original integral as:

∫(A/(x - 1) + B/(x - 2) + C/(x + 2)) dx

Now, we can integrate each term separately to find the final result.

1. Factor the denominator:

$$x^2 - 4 = (x - 2)(x + 2)$$

2. Perform partial fraction decomposition:

$$\frac{x^2 - 2x - 3}{(x - 1)(x- 2)(x + 2)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 - 4}$$

Multiplying by the denominator and equating coefficients:

- For x^2 term: A + B = 1
- For x term: -A + C = -2
- For constant term: -3 = -A

3. Solve for A, B, and C:

- From the third equation, A = 3
- From the first equation, B = 1 - 3 = -2
- From the second equation, C = -2 - 3 = -5

4. Rewrite the integrand:

$$\frac{x^2 - 2x - 3}{(x - 1)(x^2 - 4)} = \frac{3}{x - 1} - \frac{2x + 5}{x^2 - 4}$$

5. Integrate:

$$\int\frac{x^2 - 2x - 3}{(x - 1)(x^2 - 4)}dx = \int\frac{3}{x - 1}dx - \int\frac{2x + 5}{x^2 - 4}dx$$

$$= 3\ln|x - 1| - \frac{1}{2}\ln|x^2 - 4| + C$$