Answer :
To solve the integral [tex]\(\int \frac{2x + 3}{x^2 - 3x + 2} \, dx\)[/tex], we can use partial fraction decomposition since the integrand is a rational function.
### Step 1: Factor the Denominator
The first step is to factor the quadratic polynomial in the denominator:
[tex]\[ x^2 - 3x + 2 \][/tex]
This can be factored as:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
### Step 2: Decompose the Fraction
Next, express [tex]\(\frac{2x + 3}{(x - 1)(x - 2)}\)[/tex] as a sum of partial fractions:
[tex]\[ \frac{2x + 3}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants to be determined.
### Step 3: Solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we write:
[tex]\[ 2x + 3 = A(x - 2) + B(x - 1) \][/tex]
Expanding and collecting like terms, we get:
[tex]\[ 2x + 3 = Ax - 2A + Bx - B \][/tex]
[tex]\[ 2x + 3 = (A + B)x - (2A + B) \][/tex]
Comparing coefficients, we obtain two equations:
[tex]\[ A + B = 2 \][/tex]
[tex]\[ -2A - B = 3 \][/tex]
Solving this system of linear equations:
From the first equation, express [tex]\(B\)[/tex] in terms of [tex]\(A\)[/tex]:
[tex]\[ B = 2 - A \][/tex]
Substitute [tex]\(B = 2 - A\)[/tex] into the second equation:
[tex]\[ -2A - (2 - A) = 3 \][/tex]
[tex]\[ -2A - 2 + A = 3 \][/tex]
[tex]\[ -A - 2 = 3 \][/tex]
[tex]\[ -A = 5 \][/tex]
[tex]\[ A = -5 \][/tex]
Now, substitute [tex]\(A = -5\)[/tex] back into [tex]\(B = 2 - A\)[/tex]:
[tex]\[ B = 2 - (-5) \][/tex]
[tex]\[ B = 7 \][/tex]
### Step 4: Write the Integral in Terms of Partial Fractions
With [tex]\(A\)[/tex] and [tex]\(B\)[/tex] found, the partial fraction decomposition is:
[tex]\[ \frac{2x + 3}{(x - 1)(x - 2)} = \frac{-5}{x - 1} + \frac{7}{x - 2} \][/tex]
Thus, the integral becomes:
[tex]\[ \int \frac{2x + 3}{x^2 - 3x + 2} \, dx = \int \left( \frac{-5}{x - 1} + \frac{7}{x - 2} \right) \, dx \][/tex]
### Step 5: Integrate Each Term
Integrate each term separately:
[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]
[tex]\[ = -5 \ln|x - 1| + 7 \ln|x - 2| + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
Therefore, the indefinite integral is:
[tex]\[ \int \frac{2x + 3}{x^2 - 3x + 2} \, dx = 7 \ln|x - 2| - 5 \ln|x - 1| + C \][/tex]
### Step 1: Factor the Denominator
The first step is to factor the quadratic polynomial in the denominator:
[tex]\[ x^2 - 3x + 2 \][/tex]
This can be factored as:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
### Step 2: Decompose the Fraction
Next, express [tex]\(\frac{2x + 3}{(x - 1)(x - 2)}\)[/tex] as a sum of partial fractions:
[tex]\[ \frac{2x + 3}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants to be determined.
### Step 3: Solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we write:
[tex]\[ 2x + 3 = A(x - 2) + B(x - 1) \][/tex]
Expanding and collecting like terms, we get:
[tex]\[ 2x + 3 = Ax - 2A + Bx - B \][/tex]
[tex]\[ 2x + 3 = (A + B)x - (2A + B) \][/tex]
Comparing coefficients, we obtain two equations:
[tex]\[ A + B = 2 \][/tex]
[tex]\[ -2A - B = 3 \][/tex]
Solving this system of linear equations:
From the first equation, express [tex]\(B\)[/tex] in terms of [tex]\(A\)[/tex]:
[tex]\[ B = 2 - A \][/tex]
Substitute [tex]\(B = 2 - A\)[/tex] into the second equation:
[tex]\[ -2A - (2 - A) = 3 \][/tex]
[tex]\[ -2A - 2 + A = 3 \][/tex]
[tex]\[ -A - 2 = 3 \][/tex]
[tex]\[ -A = 5 \][/tex]
[tex]\[ A = -5 \][/tex]
Now, substitute [tex]\(A = -5\)[/tex] back into [tex]\(B = 2 - A\)[/tex]:
[tex]\[ B = 2 - (-5) \][/tex]
[tex]\[ B = 7 \][/tex]
### Step 4: Write the Integral in Terms of Partial Fractions
With [tex]\(A\)[/tex] and [tex]\(B\)[/tex] found, the partial fraction decomposition is:
[tex]\[ \frac{2x + 3}{(x - 1)(x - 2)} = \frac{-5}{x - 1} + \frac{7}{x - 2} \][/tex]
Thus, the integral becomes:
[tex]\[ \int \frac{2x + 3}{x^2 - 3x + 2} \, dx = \int \left( \frac{-5}{x - 1} + \frac{7}{x - 2} \right) \, dx \][/tex]
### Step 5: Integrate Each Term
Integrate each term separately:
[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]
[tex]\[ = -5 \ln|x - 1| + 7 \ln|x - 2| + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
Therefore, the indefinite integral is:
[tex]\[ \int \frac{2x + 3}{x^2 - 3x + 2} \, dx = 7 \ln|x - 2| - 5 \ln|x - 1| + C \][/tex]
