To solve the integral [tex]\(\int \frac{2x + 3}{x^2 - 3x + 2} \, dx\)[/tex], let's break it down step by step.
1. Factor the denominator:
The denominator [tex]\(x^2 - 3x + 2\)[/tex] can be factored. We need to find the roots of the quadratic equation [tex]\(x^2 - 3x + 2 = 0\)[/tex]:
[tex]\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\][/tex]
So, we rewrite the integral as:
[tex]\[
\int \frac{2x + 3}{(x - 1)(x - 2)} \, dx
\][/tex]
2. Perform partial fraction decomposition:
We express [tex]\(\frac{2x + 3}{(x - 1)(x - 2)}\)[/tex] as the sum of partial fractions:
[tex]\[
\frac{2x + 3}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}
\][/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we clear the denominators by multiplying through by [tex]\((x - 1)(x - 2)\)[/tex]:
[tex]\[
2x + 3 = A(x - 2) + B(x - 1)
\][/tex]
Expand and simplify:
[tex]\[
2x + 3 = Ax - 2A + Bx - B
\][/tex]
Combine like terms:
[tex]\[
2x + 3 = (A + B)x - 2A - B
\][/tex]
This gives us a system of equations:
[tex]\[
A + B = 2 \quad \text{(coefficient of \(x\))}
\][/tex]
[tex]\[
-2A - B = 3 \quad \text{(constant term)}
\][/tex]
3. Solve the system of equations:
From the first equation:
[tex]\[
B = 2 - A
\][/tex]
Substitute [tex]\(B = 2 - A\)[/tex] into the second equation:
[tex]\[
-2A - (2 - A) = 3
\][/tex]
Simplify:
[tex]\[
-2A - 2 + A = 3
\][/tex]
[tex]\[
-A - 2 = 3
\][/tex]
[tex]\[
-A = 5
\][/tex]
[tex]\[
A = -5
\][/tex]
Substitute [tex]\(A = -5\)[/tex] back into [tex]\(B = 2 - A\)[/tex]:
[tex]\[
B = 2 - (-5)
\][/tex]
[tex]\[
B = 7
\][/tex]
4. Rewrite the integral in terms of partial fractions:
[tex]\[
\int \frac{2x + 3}{(x - 1)(x - 2)} \, dx = \int \left( \frac{-5}{x - 1} + \frac{7}{x - 2} \right) \, dx
\][/tex]
5. Integrate term by term:
[tex]\[
\int \left( \frac{-5}{x - 1} + \frac{7}{x - 2} \right) \, dx = -5 \int \frac{1}{x - 1} \, dx + 7 \int \frac{1}{x - 2} \, dx
\][/tex]
The integral of [tex]\(\frac{1}{u}\, du\)[/tex] is [tex]\(\ln|u|\)[/tex], so we get:
[tex]\[
-5 \ln|x - 1| + 7 \ln|x - 2| + C
\][/tex]
6. Combine the logarithms (omitting absolute value notation for simplicity):
[tex]\[
7 \ln(x - 2) - 5 \ln(x - 1) + C
\][/tex]
Thus, the final result of the integral is:
[tex]\[
\boxed{7 \ln(x - 2) - 5 \ln(x - 1) + C}
\][/tex]
