Let's solve the integral:

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

Step 1: Factor the denominator

First, factor the quadratic expression in the denominator:

$x^2 + 4x + 3 = (x+1)(x+3)$

So, the integral becomes:

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

Step 2: Partial Fraction Decomposition

We can now decompose the fraction into simpler terms using partial fractions. We express:

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

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

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

Step 3: Solve for $A$ and $B$

Expand both sides:

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

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

• Coefficient of $x$: $A + B = 1$
• Constant term: $3A + B = -3$

Solve the system of equations:

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

Subtract the first equation from the second:

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

Substitute $A = -2$ into $A + B = 1$:

$-2 + B = 1$ $B = 3$

Step 4: Write the integral

Now, substitute $A$ and $B$ into the partial fractions:

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

So, the integral becomes:

$\int \frac{-2}{x+1} \, dx + \int \frac{3}{x+3} \, dx$

Step 5: Integrate

Now integrate each term:

$\int \frac{-2}{x+1} \, dx = -2 \ln |x+1|$ $\int \frac{3}{x+3} \, dx = 3 \ln |x+3|$

Step 6: Combine the results

The integral is:

$-2 \ln |x+1| + 3 \ln |x+3| + C$

Where $C$ is the constant of integration.

Final Answer:

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

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Would you like more details on any step or have any questions about the solution? Here are some related questions you might find interesting:

1. How do you apply partial fraction decomposition to more complex rational functions?
2. What are the different methods for solving integrals involving rational functions?
3. How do you check the correctness of a partial fraction decomposition?
4. Can the result be simplified further in any way?
5. What other techniques can be used to handle integrals involving quadratics in the denominator?

Tip: Always check if the denominator can be factored when dealing with rational functions, as this often simplifies the process of partial fraction decomposition.