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

To solve the given problem, we'll perform the following steps: 1. **Decompose the function \( f(x) \) into partial fractions.** 2. **Evaluate the definite integral of \( f(x) \) from 2 to 3.** Let's proceed step-by-step. --- ### 1. Partial Fraction Decomposition Given the function: \[ f(x) = \frac{x + 2}{(x^2 + 1)(2x - 1)} \] We aim to express \( f(x) \) as a sum of simpler fractions: \[ \frac{x + 2}{(x^2 + 1)(2x - 1)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{2x - 1} \] **Step-by-Step Decomposition:** 1. **Set Up the Equation:** \[ \frac{x + 2}{(x^2 + 1)(2x - 1)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{2x - 1} \] 2. **Multiply Both Sides by \((x^2 + 1)(2x - 1)\) to Clear the Denominator:** \[ x + 2 = (Ax + B)(2x - 1) + C(x^2 + 1) \] 3. **Expand the Right Side:** \[ x + 2 = 2Ax^2 - Ax + 2Bx - B + Cx^2 + C \] 4. **Combine Like Terms:** \[ x + 2 = (2A + C)x^2 + (-A + 2B)x + (-B + C) \] 5. **Set Up a System of Equations by Equating Coefficients:** \[ \begin{cases} 2A + C = 0 \quad &\text{(Coefficient of } x^2\text{)} \\ -A + 2B = 1 \quad &\text{(Coefficient of } x\text{)} \\ -B + C = 2 \quad &\text{(Constant term)} \end{cases} \] 6. **Solve the System:** - From the first equation: \( C = -2A \) - Substitute \( C \) into the third equation: \( -B - 2A = 2 \) → \( B + 2A = -2 \) - Substitute \( B = -2 - 2A \) into the second equation: \( -A + 2(-2 - 2A) = 1 \) → \( -A - 4 - 4A = 1 \) → \( -5A = 5 \) → \( A = -1 \) - Then, \( B = -2 - 2(-1) = 0 \) - Finally, \( C = -2(-1) = 2 \) 7. **Write the Partial Fractions:** \[ f(x) = \frac{-x}{x^2 + 1} + \frac{2}{2x - 1} \] --- ### 2. Evaluating the Definite Integral We now compute: \[ \int_{2}^{3} f(x)\,dx = \int_{2}^{3} \left( \frac{-x}{x^2 + 1} + \frac{2}{2x - 1} \right) dx \] **Step-by-Step Integration:** 1. **Separate the Integral:** \[ \int \frac{-x}{x^2 + 1} \, dx + \int \frac{2}{2x - 1} \, dx \] 2. **Integrate Each Term Individually:** - **First Integral:** \[ \int \frac{-x}{x^2 + 1} \, dx \] Let \( u = x^2 + 1 \), then \( du = 2x\,dx \). \[ \int \frac{-x}{x^2 + 1} \, dx = -\frac{1}{2} \ln|x^2 + 1| + C \] - **Second Integral:** \[ \int \frac{2}{2x - 1} \, dx \] Let \( u = 2x - 1 \), then \( du = 2\,dx \), so \( dx = \frac{du}{2} \). \[ \int \frac{2}{2x - 1} \, dx = \ln|2x - 1| + C \] 3. **Combine the Results:** \[ \int f(x)\,dx = -\frac{1}{2} \ln(x^2 + 1) + \ln(2x - 1) + C \] 4. **Evaluate from 2 to 3:** \[ \left[ -\frac{1}{2} \ln(x^2 + 1) + \ln(2x - 1) \right]_{2}^{3} \] \[ = \left( -\frac{1}{2} \ln(10) + \ln(5) \right) - \left( -\frac{1}{2} \ln(5) + \ln(3) \right) \] \[ = -\frac{1}{2} \ln(10) + \ln(5) + \frac{1}{2} \ln(5) - \ln(3) \] \[ = -\frac{1}{2} \ln(10) + \frac{3}{2} \ln(5) - \ln(3) \] 5. **Calculate Numerically:** Using natural logarithm values: \[ \ln(10) \approx 2.302585093,\quad \ln(5) \approx 1.609437912,\quad \ln(3) \approx 1.098612289 \] \[ -\frac{1}{2} \times 2.302585093 \approx -1.151292547 \] \[ \frac{3}{2} \times 1.609437912 \approx 2.414156868 \] \[ -1.098612289 \] Combine all: \[ -1.151292547 + 2.414156868 - 1.098612289 \approx 0.164252032 \] Rounding to four decimal places: \[ \int_{2}^{3} f(x)\,dx \approx 0.1643 \] --- ### **Final Answer:** After decomposition, \[ f(x) = \frac{-x}{x² + 1} + \frac{2}{2x - 1} \] and the definite integral from 2 to 3 is approximately 0.1643.