\( \frac{1}{x-2}+\frac{2}{x-3} \) actions \( \int=\frac{1}{(x-2) \times(x-3)}+\frac{2 \times(x-2)}{(x-3) \times(x-2)} \)

To solve the expression \( \frac{1}{x-2}+\frac{2}{x-3} \), we will first find a common denominator and then integrate the resulting expression. ### Step 1: Find a common denominator The common denominator for the fractions \( \frac{1}{x-2} \) and \( \frac{2}{x-3} \) is \( (x-2)(x-3) \). ### Step 2: Rewrite the fractions We can rewrite the fractions with the common denominator: \[ \frac{1}{x-2} = \frac{(x-3)}{(x-2)(x-3)} \] \[ \frac{2}{x-3} = \frac{2(x-2)}{(x-3)(x-2)} \] ### Step 3: Combine the fractions Now we can combine the two fractions: \[ \frac{(x-3) + 2(x-2)}{(x-2)(x-3)} \] ### Step 4: Simplify the numerator Now, let's simplify the numerator: \[ (x-3) + 2(x-2) = x - 3 + 2x - 4 = 3x - 7 \] So, we have: \[ \frac{3x - 7}{(x-2)(x-3)} \] ### Step 5: Integrate the expression Now we need to integrate: \[ \int \frac{3x - 7}{(x-2)(x-3)} \, dx \] To perform the integration, we can use partial fraction decomposition. We will express \( \frac{3x - 7}{(x-2)(x-3)} \) as: \[ \frac{A}{x-2} + \frac{B}{x-3} \] Multiplying through by the denominator \( (x-2)(x-3) \) gives: \[ 3x - 7 = A(x-3) + B(x-2) \] Now, we will solve for \( A \) and \( B \). ### Step 6: Set up the equations Expanding the right side: \[ 3x - 7 = Ax - 3A + Bx - 2B = (A + B)x - (3A + 2B) \] This gives us the system of equations: 1. \( A + B = 3 \) 2. \( -3A - 2B = -7 \) Now, let's solve this system of equations. ### Step 7: Solve the system of equations We can solve these equations simultaneously. 1. From the first equation, we can express \( B \) in terms of \( A \): \[ B = 3 - A \] 2. Substitute \( B \) into the second equation: \[ -3A - 2(3 - A) = -7 \] This simplifies to: \[ -3A - 6 + 2A = -7 \] Combining like terms gives: \[ -A - 6 = -7 \] So, \[ -A = -1 \implies A = 1 \] Substituting \( A = 1 \) back into the first equation: \[ 1 + B = 3 \implies B = 2 \] ### Step 8: Write the partial fractions Now we have: \[ \frac{3x - 7}{(x-2)(x-3)} = \frac{1}{x-2} + \frac{2}{x-3} \] ### Step 9: Integrate each term Now we can integrate: \[ \int \left( \frac{1}{x-2} + \frac{2}{x-3} \right) \, dx = \int \frac{1}{x-2} \, dx + 2 \int \frac{1}{x-3} \, dx \] The integrals are: \[ \int \frac{1}{x-2} \, dx = \ln |x-2| + C_1 \] \[ \int \frac{1}{x-3} \, dx = \ln |x-3| + C_2 \] Thus, the final result is: \[ \int \frac{3x - 7}{(x-2)(x-3)} \, dx = \ln |x-2| + 2\ln |x-3| + C \] where \( C \) is the constant of integration.