Evaluate using partial fractions \( I=\int_{1}^{\infty} \frac{x-1}{(x+1)\left(x^{2}+1\right)} d x \)

To evaluate the integral \[ I = \int_{1}^{\infty} \frac{x-1}{(x+1)(x^2+1)} \, dx, \] we will first decompose the integrand into partial fractions. We can express \[ \frac{x-1}{(x+1)(x^2+1)} \] in the form \[ \frac{A}{x+1} + \frac{Bx + C}{x^2 + 1}. \] Multiplying both sides by the denominator \((x+1)(x^2+1)\) gives: \[ x - 1 = A(x^2 + 1) + (Bx + C)(x + 1). \] Expanding the right-hand side: \[ x - 1 = A(x^2 + 1) + Bx^2 + Bx + Cx + C. \] Combining like terms, we have: \[ x - 1 = (A + B)x^2 + (B + C)x + (A + C). \] Now, we can equate coefficients from both sides: 1. For \(x^2\): \(A + B = 0\) 2. For \(x\): \(B + C = 1\) 3. For the constant term: \(A + C = -1\) From the first equation, we can express \(B\) in terms of \(A\): \[ B = -A. \] Substituting \(B = -A\) into the second equation: \[ -A + C = 1 \implies C = A + 1. \] Now substituting \(B = -A\) and \(C = A + 1\) into the third equation: \[ A + (A + 1) = -1 \implies 2A + 1 = -1 \implies 2A = -2 \implies A = -1. \] Now substituting \(A = -1\) back to find \(B\) and \(C\): \[ B = -(-1) = 1, \] \[ C = -1 + 1 = 0. \] Thus, we have: \[ \frac{x-1}{(x+1)(x^2+1)} = \frac{-1}{x+1} + \frac{x}{x^2+1}. \] Now we can rewrite the integral \(I\): \[ I = \int_{1}^{\infty} \left( \frac{-1}{x+1} + \frac{x}{x^2+1} \right) \, dx. \] We can split this into two separate integrals: \[ I = -\int_{1}^{\infty} \frac{1}{x+1} \, dx + \int_{1}^{\infty} \frac{x}{x^2+1} \, dx. \] **Evaluating the first integral:** \[ -\int_{1}^{\infty} \frac{1}{x+1} \, dx = -\left[ \ln(x+1) \right]_{1}^{\infty} = -\left( \lim_{b \to \infty} \ln(b+1) - \ln(2) \right) = -\infty. \] This integral diverges. **Evaluating the second integral:** For the second integral, we can use the substitution \(u = x^2 + 1\), \(du = 2x \, dx\), or \(dx = \frac{du}{2\sqrt{u-1}}\): \[ \int \frac{x}{x^2+1} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{u-1}} \, du. \] However, we can also evaluate it directly: \[ \int_{1}^{\infty} \frac{x}{x^2+1} \, dx = \left[ \frac{1}{2} \ln(x^2 + 1) \right]_{1}^{\infty} = \frac{1}{2} \left( \lim_{b \to \infty} \ln(b^2 + 1) - \ln(2) \right) = \infty. \] Thus, both integrals diverge. Since both parts diverge, we conclude that the original integral \(I\) diverges: \[ \boxed{\infty}. \]