determine a integral de (x^2-x-4)/(x^2-2x-15)

To find the integral of the expression \(\frac{x^2 - x - 4}{x^2 - 2x - 15}\), we'll first simplify the integrand by performing polynomial long division, if necessary, and then look for a way to integrate the resulting expression.

### 1. **Simplify the integrand:**

The given expression is:

\[
\frac{x^2 - x - 4}{x^2 - 2x - 15}
\]

First, let's factor the denominator \(x^2 - 2x - 15\):

\[
x^2 - 2x - 15 = (x - 5)(x + 3)
\]

So, the integral becomes:

\[
\int \frac{x^2 - x - 4}{(x - 5)(x + 3)} \, dx
\]

### 2. **Polynomial Long Division:**

Since the degree of the numerator \(x^2 - x - 4\) is equal to the degree of the denominator \(x^2 - 2x - 15\), we can perform polynomial long division:

Divide \(x^2 - x - 4\) by \(x^2 - 2x - 15\):

\[
\text{Quotient: } 1 \quad \text{Remainder: } x
\]

This gives us:

\[
\frac{x^2 - x - 4}{x^2 - 2x - 15} = 1 + \frac{(x + 2)}{(x - 5)(x + 3)}
\]

So, the integral simplifies to:

\[
\int \frac{x^2 - x - 4}{x^2 - 2x - 15} \, dx = \int 1 \, dx + \int \frac{x + 2}{(x - 5)(x + 3)} \, dx
\]

### 3. **Integrate the simplified expression:**

The first part is straightforward:

\[
\int 1 \, dx = x
\]

For the second part, we use partial fraction decomposition.

### 4. **Partial Fraction Decomposition:**

We express:

\[
\frac{x + 2}{(x - 5)(x + 3)} = \frac{A}{x - 5} + \frac{B}{x + 3}
\]

Multiply through by the denominator to clear the fractions:

\[
x + 2 = A(x + 3) + B(x - 5)
\]

Expanding and equating coefficients:

\[
x + 2 = A(x + 3) + B(x - 5) = (A + B)x + (3A - 5B)
\]

This gives us two equations:

\[
A + B = 1 \quad \text{(coefficient of } x \text{)}
\]
\[
3A - 5B = 2 \quad \text{(constant term)}
\]

Solving this system of equations:

1. \( A + B = 1 \)
2. \( 3A - 5B = 2 \)

From the first equation, \(A = 1 - B\). Substitute into the second equation:

\[
3(1 - B) - 5B = 2
\]

\[
3 - 3B - 5B = 2
\]

\[
-8B = -1 \quad \Rightarrow \quad B = \frac{1}{8}
\]

Substitute \(B = \frac{1}{8}\) back into \(A + B = 1\):

\[
A = 1 - \frac{1}{8} = \frac{7}{8}
\]

So, the partial fractions are:

\[
\frac{x + 2}{(x - 5)(x + 3)} = \frac{7/8}{x - 5} + \frac{1/8}{x + 3}
\]

### 5. **Integrate using the partial fractions:**

Now, the integral becomes:

\[
\int 1 \, dx + \frac{7}{8} \int \frac{1}{x - 5} \, dx + \frac{1}{8} \int \frac{1}{x + 3} \, dx
\]

\[
= x + \frac{7}{8} \ln |x - 5| + \frac{1}{8} \ln |x + 3| + C
\]

### **Final Answer:**

The integral is:

\[
\int \frac{x^2 - x - 4}{x^2 - 2x - 15} \, dx = x + \frac{7}{8} \ln |x - 5| + \frac{1}{8} \ln |x + 3| + C
\]

Would you like more details or have any questions?

### Related Questions:
1. How do you factor quadratic polynomials?
2. What is the method of partial fractions, and how is it used in integration?
3. How do you perform polynomial long division?
4. How do you integrate functions of the form \(\frac{1}{ax + b}\)?
5. Can you solve integrals involving trigonometric functions?

**Tip:** When integrating rational functions, always consider polynomial long division first if the degree of the numerator is greater than or equal to the degree of the denominator.

Learn how to integrate the rational function (x^2 - x - 4)/(x^2 - 2x - 15) using polynomial long division and partial fraction decomposition. This detailed solution covers integral calculus techniques suitable for college-level students.

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