The integral you're dealing with is:

$\int \frac{x - 1}{x^2 - 5x - 6} \, dx$

The denominator $x^2 - 5x - 6$ can be factored to simplify the expression. Let's first factor the quadratic:

$x^2 - 5x - 6 = (x - 6)(x + 1)$

So the integral becomes:

$\int \frac{x - 1}{(x - 6)(x + 1)} \, dx$

Next, we'll use partial fraction decomposition. We'll express the fraction as:

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

Multiplying both sides by $(x - 6)(x + 1)$ and solving for $A$ and $B$:

$x - 1 = A(x + 1) + B(x - 6)$

Now, expand both sides:

$x - 1 = A(x) + A + B(x) - 6B$

Combine like terms:

$x - 1 = (A + B)x + (A - 6B)$

By comparing coefficients of $x$ and the constant term, we get two equations:

1. $A + B = 1$
2. $A - 6B = -1$

Solving this system:

From $A + B = 1$, we have $A = 1 - B$. Substitute this into the second equation:

$(1 - B) - 6B = -1$

Simplifying:

$1 - B - 6B = -1$

$1 - 7B = -1$

$-7B = -2 \implies B = \frac{2}{7}$

Now substitute $B = \frac{2}{7}$ into $A + B = 1$:

$A + \frac{2}{7} = 1 \implies A = 1 - \frac{2}{7} = \frac{5}{7}$

Thus, the partial fraction decomposition is:

$\frac{x - 1}{(x - 6)(x + 1)} = \frac{5}{7(x - 6)} + \frac{2}{7(x + 1)}$

Now, we can integrate each term:

$\int \frac{5}{7(x - 6)} \, dx + \int \frac{2}{7(x + 1)} \, dx$

The integrals are simple logarithmic forms:

$= \frac{5}{7} \ln|x - 6| + \frac{2}{7} \ln|x + 1| + C$

Thus, the solution to the integral is:

$\boxed{\frac{5}{7} \ln|x - 6| + \frac{2}{7} \ln|x + 1| + C}$

Would you like a detailed explanation of any steps, or do you have any questions?

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Here are 5 related questions:

1. How do you perform partial fraction decomposition in general?
2. Can you explain when to use logarithmic integration?
3. What is the general rule for integrating rational functions?
4. How do you solve systems of equations with two variables?
5. Can this method be extended to higher-order polynomials?

Tip: Always look for factorizations in the denominator when solving integrals of rational functions.