Use partial fractions to evaluate the definite integral. \[ \int_{1}^{3} \frac{x+1}{x\left(x^{2}+1\right)} d x \]

To evaluate the definite integral \[ \int_{1}^{3} \frac{x+1}{x(x^{2}+1)} \, dx, \] we first decompose the integrand into partial fractions. We want to express \[ \frac{x+1}{x(x^{2}+1)} = \frac{A}{x} + \frac{Bx+C}{x^{2}+1}, \] where \( A \), \( B \), and \( C \) are constants to be determined. Multiplying through by the denominator \( x(x^{2}+1) \) gives: \[ x + 1 = A(x^{2}+1) + (Bx+C)x. \] Expanding the right-hand side, we have: \[ x + 1 = Ax^{2} + A + Bx^{2} + Cx. \] Combining like terms results in: \[ x + 1 = (A + B)x^{2} + Cx + A. \] To find the coefficients, we equate coefficients from both sides. This gives us the following system of equations: 1. \( A + B = 0 \) (coefficient of \( x^{2} \)) 2. \( C = 1 \) (coefficient of \( x \)) 3. \( A = 1 \) (constant term) From equation 3, we have \( A = 1 \). Substituting into equation 1, we find \( B = -1 \). From equation 2, we already have \( C = 1 \). Thus, we have: \[ A = 1, \quad B = -1, \quad C = 1. \] This gives the partial fraction decomposition: \[ \frac{x+1}{x(x^{2}+1)} = \frac{1}{x} + \frac{-x + 1}{x^{2}+1}. \] So the integral becomes: \[ \int_{1}^{3} \left( \frac{1}{x} + \frac{-x + 1}{x^{2}+1} \right) \, dx. \] We split the integral into two parts: \[ \int_{1}^{3} \frac{1}{x} \, dx + \int_{1}^{3} \frac{-x + 1}{x^{2}+1} \, dx. \] Calculating the first integral: \[ \int_{1}^{3} \frac{1}{x} \, dx = \left[ \ln |x| \right]_{1}^{3} = \ln 3 - \ln 1 = \ln 3. \] For the second integral, we split it as well: \[ \int_{1}^{3} \frac{-x}{x^{2}+1} \, dx + \int_{1}^{3} \frac{1}{x^{2}+1} \, dx. \] The first part can be evaluated using substitution. Let \( u = x^{2}+1 \), then \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \). When \( x = 1 \), \( u = 2 \) and when \( x = 3 \), \( u = 10 \). Thus, \[ \int_{1}^{3} \frac{-x}{x^{2}+1} \, dx = -\frac{1}{2} \int_{2}^{10} \frac{1}{u} \, du = -\frac{1}{2} \left[ \ln u \right]_{2}^{10} = -\frac{1}{2}(\ln 10 - \ln 2) = -\frac{1}{2} \ln \frac{10}{2} = -\frac{1}{2} \ln 5. \] The second part is straightforward: \[ \int_{1}^{3} \frac{1}{x^{2}+1} \, dx = \left[ \tan^{-1} x \right]_{1}^{3} = \tan^{-1}(3) - \tan^{-1}(1) = \tan^{-1}(3) - \frac{\pi}{4}. \] Putting it all together, the integral becomes: \[ \int_{1}^{3} \frac{x+1}{x(x^{2}+1)} \, dx = \ln 3 - \left( -\frac{1}{2} \ln 5 + \tan^{-1}(3) - \frac{\pi}{4} \right). \] Therefore, the final result is: \[