rallel to AC meets BC at E . Find the co-ordinates of point E . press the function \( f(x)=\frac{x+2}{\left(x^{2}+1\right)(2 x-1)} \) as a sum of partial fractions. Hence find \( \int_{2}^{3} f(x) d x \), co 4 decimal places

Certainly! Let's address the two parts of your query: 1. **Finding the Coordinates of Point E:** Unfortunately, the first part of your question seems incomplete as it references points and a diagram that aren't provided. To determine the coordinates of point \( E \), more information or a diagram is needed, such as the coordinates of points \( A \), \( B \), and \( C \), and the specific condition for the line parallel to \( AC \). *If you can provide additional details or clarify the problem, I'd be happy to assist further!* 2. **Expressing \( f(x) = \frac{x+2}{(x^{2}+1)(2x-1)} \) as Partial Fractions and Evaluating the Integral:** **a. Partial Fraction Decomposition:** We aim to express \( f(x) \) as a sum of simpler fractions: \[ \frac{x+2}{(x^{2}+1)(2x-1)} = \frac{Ax + B}{x^{2}+1} + \frac{C}{2x -1} \] **Steps:** 1. **Multiply both sides by \((x^{2}+1)(2x-1)\):** \[ x + 2 = (Ax + B)(2x -1) + C(x^{2} +1) \] 2. **Expand and simplify:** \[ x + 2 = 2A x^{2} - A x + 2B x - B + C x^{2} + C \] \[ x + 2 = (2A + C) x^{2} + (-A + 2B) x + (-B + C) \] 3. **Equate the coefficients of like terms:** \[ \begin{cases} 2A + C = 0 \quad &(1) \\ -A + 2B = 1 \quad &(2) \\ -B + C = 2 \quad &(3) \end{cases} \] 4. **Solve the system of equations:** From equation (1): \( C = -2A \) Substitute \( C \) into equation (3): \[ -B - 2A = 2 \quad \Rightarrow \quad B = -2A - 2 \] Substitute \( B \) into equation (2): \[ -A + 2(-2A -2) = 1 \quad \Rightarrow \quad -A -4A -4 = 1 \quad \Rightarrow \quad -5A = 5 \quad \Rightarrow \quad A = -1 \] Then: \[ C = -2(-1) = 2 \] \[ B = -2(-1) - 2 = 0 \] 5. **Final Partial Fractions:** \[ f(x) = \frac{-x}{x^{2}+1} + \frac{2}{2x -1} \] **b. Evaluating the Integral:** We need to compute: \[ \int_{2}^{3} f(x) \, dx = \int_{2}^{3} \left( -\frac{x}{x^{2}+1} + \frac{2}{2x -1} \right) dx \] **Steps:** 1. **Integrate term by term:** \[ \int \left( -\frac{x}{x^{2}+1} \right) dx = -\frac{1}{2} \ln(x^{2} + 1) + C \] \[ \int \frac{2}{2x -1} dx = \ln|2x - 1| + C \] 2. **Combine the results:** \[ \int_{2}^{3} f(x) \, dx = \left[ -\frac{1}{2} \ln(x^{2} + 1) + \ln(2x -1) \right]_{2}^{3} \] 3. **Evaluate the definite integral:** \[ = \left( -\frac{1}{2} \ln(10) + \ln(5) \right) - \left( -\frac{1}{2} \ln(5) + \ln(3) \right) \] \[ = -\frac{1}{2} \ln(10) + \frac{3}{2} \ln(5) - \ln(3) \] 4. **Calculate numerically:** Using approximate natural logarithm values: \[ \ln(10) \approx 2.302585093 \] \[ \ln(5) \approx 1.609437912 \] \[ \ln(3) \approx 1.098612289 \] Substitute these values: \[ -\frac{1}{2} \times 2.302585093 \approx -1.151292547 \] \[ \frac{3}{2} \times 1.609437912 \approx 2.414156868 \] \[ -1.098612289 \] Combine them: \[ -1.151292547 + 2.414156868 - 1.098612289 \approx 0.164252032 \] 5. **Final Answer (to 4 decimal places):** \[ \int_{2}^{3} f(x) \, dx \approx 0.1643 \] **Summary:** \[ \int_{2}^{3} \frac{x+2}{(x^{2}+1)(2x-1)} \, dx \approx 0.1643 \]