\( \int_{1}^{\text {(d) }} \int_{1}^{3} \frac{4 x+1}{2 x^{2}+x} d x \)

To solve the double integral \( \int_{1}^{\text {(d) }} \int_{1}^{3} \frac{4 x + 1}{2 x^2 + x} \, dx \), we first need to focus on the inner integral \( \int_{1}^{3} \frac{4 x + 1}{2 x^2 + x} \, dx \). ### Step 1: Simplifying the Inner Integral We can simplify the integrand \( \frac{4 x + 1}{2 x^2 + x} \). Start by factoring the denominator: \[ 2 x^2 + x = x(2x + 1) \] Now let's perform polynomial long division or recognize the structure if possible. We can separate the integrand: \[ \frac{4 x + 1}{x(2x + 1)} = \frac{A}{x} + \frac{B}{2x + 1} \] Multiplying through by the denominator \( x(2x + 1) \) gives: \[ 4x + 1 = A(2x + 1) + Bx \] Expanding gives: \[ 4x + 1 = (2A + B)x + A \] This leads to the system of equations: 1. \( 2A + B = 4 \) 2. \( A = 1 \) From \( A = 1 \), we substitute into the first equation: \[ 2(1) + B = 4 \implies B = 2 \] So we can rewrite the integrand as: \[ \frac{4x + 1}{2x^2 + x} = \frac{1}{x} + \frac{2}{2x + 1} \] ### Step 2: Evaluating the Inner Integral Now we can find the inner integral: \[ \int_{1}^{3} \left( \frac{1}{x} + \frac{2}{2x + 1} \right) dx = \int_{1}^{3} \frac{1}{x} \, dx + 2 \int_{1}^{3} \frac{1}{2x + 1} \, dx \] 1. **The first part**: \[ \int \frac{1}{x} \, dx = \ln|x| \Rightarrow \left[ \ln x \right]_{1}^{3} = \ln 3 - \ln 1 = \ln 3 \] 2. **The second part**: Using the substitution \( u = 2x + 1 \), hence \( du = 2dx \Rightarrow dx = \frac{du}{2} \). The limits change as follows: when \( x = 1, u = 3 \) and when \( x = 3, u = 7 \). \[ 2 \int_{1}^{3} \frac{1}{2x + 1} \, dx = 2 \cdot \frac{1}{2} \int_{3}^{7} \frac{1}{u} \, du = \int_{3}^{7} \frac{1}{u} \, du = \left[ \ln u \right]_{3}^{7} = \ln 7 - \ln 3 = \ln \frac{7}{3} \] ### Combining the Results Now, summing these results gives us: \[ \int_{1}^{3} \frac{4x + 1}{2x^2 + x} \, dx = \ln 3 + \ln \frac{7}{3} = \ln 7 \] ### Step 3: Final Integration Now substituting the result back into the outer integral: \[ \int_{1}^{d} \ln 7 \, dy \] This evaluates to: \[ \ln 7 \cdot (d - 1) \] This final expression gives us the result of the double integral: \[ \int_{1}^{d} \int_{1}^{3} \frac{4x + 1}{2x^2 + x} \, dx \, dy = \ln 7 \cdot (d - 1) \] This completes our solution! If you have a specific value for \( d \), just plug it in to find the final result. 😊