To integrate the function

$\int \frac{5x + 1}{(x + 1)(x - 3)} \, dx$

using partial fraction decomposition, we follow these steps:

Step 1: Set up the partial fraction decomposition

We want to decompose

$\frac{5x + 1}{(x + 1)(x - 3)}$

into the form

$\frac{5x + 1}{(x + 1)(x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3}$

where $A$ and $B$ are constants to be determined.

Step 2: Solve for $A$ and $B$

Multiply both sides by $(x + 1)(x - 3)$ to clear the denominators:

$5x + 1 = A(x - 3) + B(x + 1)$

Expanding the right side gives:

$5x + 1 = Ax - 3A + Bx + B$

Combine like terms:

$5x + 1 = (A + B)x + (-3A + B)$

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

1. For the $x$-terms: $A + B = 5$
2. For the constant terms: $-3A + B = 1$

Solve the system of equations:

1. $A + B = 5$
2. $-3A + B = 1$

From the first equation, we get $B = 5 - A$. Substitute this into the second equation:

$-3A + (5 - A) = 1$

Simplify:

$-4A + 5 = 1$

$-4A = -4$

$A = 1$

Now substitute $A = 1$ into the first equation:

$1 + B = 5$

$B = 4$

Thus, $A = 1$ and $B = 4$, so our partial fraction decomposition is:

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

Step 3: Integrate each term

Now we integrate:

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

These are simple logarithmic integrals:

$= \ln |x + 1| + 4 \ln |x - 3| + C$

Final Answer

The integral is:

$\int \frac{5x + 1}{(x + 1)(x - 3)} \, dx = \ln |x + 1| + 4 \ln |x - 3| + C$

where $C$ is the constant of integration.

Would you like further details on any part of the solution?

Here are five related questions to deepen your understanding:

1. How would the integral change if the numerator were $5x - 3$ instead of $5x + 1$?
2. Can you find the derivative of $\ln |x + 1| + 4 \ln |x - 3|$ to verify the result?
3. What if one of the factors in the denominator was squared? How would the partial fraction decomposition change?
4. How can you apply partial fractions to solve an improper fraction?
5. How does this integration technique extend to higher degree polynomials in the denominator?

Tip: Remember, partial fraction decomposition is particularly useful for rational functions with factors in the denominator that do not repeat.