To integrate $\frac{x+4}{x^2 - 5x + 6}$, we first simplify the expression. The denominator $x^2 - 5x + 6$ can be factored as:

$x^2 - 5x + 6 = (x - 2)(x - 3)$

So, the integral becomes:

$\int \frac{x+4}{(x-2)(x-3)} \, dx$

We will use partial fraction decomposition to rewrite the fraction. Assume:

$\frac{x+4}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$

Multiplying both sides by $(x-2)(x-3)$ to clear the denominators:

$x + 4 = A(x - 3) + B(x - 2)$

Expanding both sides:

$x + 4 = A(x) - 3A + B(x) - 2B$

$x + 4 = (A + B)x + (-3A - 2B)$

Now, equate the coefficients of $x$ and the constant terms:

1. For the $x$-terms: $A + B = 1$
2. For the constant terms: $-3A - 2B = 4$

Solving this system of equations:

1. $A + B = 1$
2. $-3A - 2B = 4$

From the first equation, we can express $B$ as $B = 1 - A$. Substituting this into the second equation:

$-3A - 2(1 - A) = 4$

$-3A - 2 + 2A = 4$

$-A - 2 = 4$

$-A = 6 \quad \Rightarrow \quad A = -6$

Substituting $A = -6$ into $A + B = 1$:

$-6 + B = 1 \quad \Rightarrow \quad B = 7$

Thus, the partial fraction decomposition is:

$\frac{x+4}{(x-2)(x-3)} = \frac{-6}{x-2} + \frac{7}{x-3}$

Now, integrate each term separately:

$\int \frac{x+4}{(x-2)(x-3)} \, dx = \int \frac{-6}{x-2} \, dx + \int \frac{7}{x-3} \, dx$

These integrals are straightforward logarithmic integrals:

$\int \frac{-6}{x-2} \, dx = -6 \ln |x-2| + C_1$

$\int \frac{7}{x-3} \, dx = 7 \ln |x-3| + C_2$

Thus, the final answer is:

$-6 \ln |x-2| + 7 \ln |x-3| + C$

where $C = C_1 + C_2$ is the constant of integration.

Would you like a detailed step on any part of the process or have any questions?

Related Questions:

1. How do you handle partial fractions when the degree of the numerator is higher than the denominator?
2. What if the denominator cannot be factored easily?
3. How does the method change when integrating rational functions with repeated roots in the denominator?
4. How do you derive logarithmic integrals like $\int \frac{1}{x} dx$?
5. What are other techniques for integrating rational functions, like using trigonometric substitutions?

Tip:

For integrals involving rational functions, always check if the denominator can be factored. Partial fraction decomposition is one of the most powerful tools for simplifying these integrals.