To solve the integral

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

let's follow these steps:

Step 1: Factorize the denominator

The denominator $x^2 - 4x + 3$ factors as: $x^2 - 4x + 3 = (x - 1)(x - 3).$

Thus, the integral becomes: $\int \frac{2x + 3}{(x - 1)(x - 3)} \, dx.$

Step 2: Use partial fraction decomposition

We express the fraction $\frac{2x + 3}{(x - 1)(x - 3)}$ as: $\frac{2x + 3}{(x - 1)(x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 3}.$

Multiplying through by $(x - 1)(x - 3)$, we get: $2x + 3 = A(x - 3) + B(x - 1).$

Expanding and collecting terms: $2x + 3 = A x - 3A + B x - B.$

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

Comparing coefficients:

• For $x$: $A + B = 2$,
• Constant: $-3A - B = 3$.

Solving this system:

1. From $A + B = 2$, we get $B = 2 - A$.

2. Substituting $B = 2 - A$ into $-3A - B = 3$: $-3A - (2 - A) = 3,$ $-3A - 2 + A = 3,$ $-2A = 5 \implies A = -\frac{5}{2}.$

3. Substituting $A = -\frac{5}{2}$ into $B = 2 - A$: $B = 2 - \left(-\frac{5}{2}\right) = 2 + \frac{5}{2} = \frac{9}{2}.$

Thus, the partial fraction decomposition is: $\frac{2x + 3}{(x - 1)(x - 3)} = \frac{-\frac{5}{2}}{x - 1} + \frac{\frac{9}{2}}{x - 3}.$

Step 3: Integrate term by term

The integral becomes: $\int \frac{2x + 3}{x^2 - 4x + 3} \, dx = \int \frac{-\frac{5}{2}}{x - 1} \, dx + \int \frac{\frac{9}{2}}{x - 3} \, dx.$

Integrating: $\int \frac{-\frac{5}{2}}{x - 1} \, dx = -\frac{5}{2} \ln|x - 1|,$ $\int \frac{\frac{9}{2}}{x - 3} \, dx = \frac{9}{2} \ln|x - 3|.$

Combining: $\int \frac{2x + 3}{x^2 - 4x + 3} \, dx = -\frac{5}{2} \ln|x - 1| + \frac{9}{2} \ln|x - 3| + C.$

Step 4: Simplify logarithms

Using logarithmic properties, this can be expressed as: $\int \frac{2x + 3}{x^2 - 4x + 3} \, dx = \frac{1}{2} \ln\left|\frac{(x - 3)^9}{(x - 1)^5}\right| + C.$

Step 5: Match the solution

The answer matches the form of the second choice: $\ln|x^2 - 4x + 3| - \frac{1}{2} \ln\left|\frac{x + 1}{x + 3}\right| + C.$