Evaluate using partial fraction \( I=\int_{1}^{\infty} \frac{x-1}{(x+1)\left(x^{2}+1\right)} d x \) A. \( \ln \frac{2}{\sqrt{2}} \) B. None of the options C. \( \ln \frac{2}{\sqrt{2}}+1 \) D. \( \ln \frac{1}{\sqrt{2}} \) E. \( \ln \sqrt{2} \)

To evaluate the integral \( I=\int_{1}^{\infty} \frac{x-1}{(x+1)\left(x^{2}+1\right)} d x \), we can use partial fraction decomposition to simplify the integral. First, let's decompose the fraction \( \frac{x-1}{(x+1)\left(x^{2}+1\right)} \) into partial fractions: \[ \frac{x-1}{(x+1)\left(x^{2}+1\right)} = \frac{A}{x+1} + \frac{Bx+C}{x^{2}+1} \] To find the values of \( A \), \( B \), and \( C \), we can equate the numerators and solve for the unknowns. \[ x-1 = A(x^{2}+1) + (Bx+C)(x+1) \] Expanding and equating coefficients, we get: \[ x-1 = Ax^{2} + A + Bx^{2} + Bx + Cx + C \] Comparing coefficients of like terms, we have: \[ A + B = 0 \] \[ A + C = -1 \] \[ B + C = 1 \] Solving this system of equations, we find: \[ A = 1, B = -1, C = 0 \] Therefore, the partial fraction decomposition is: \[ \frac{x-1}{(x+1)\left(x^{2}+1\right)} = \frac{1}{x+1} - \frac{x}{x^{2}+1} \] Now, we can integrate each term separately: \[ I = \int_{1}^{\infty} \left( \frac{1}{x+1} - \frac{x}{x^{2}+1} \right) dx \] \[ I = \int_{1}^{\infty} \frac{1}{x+1} dx - \int_{1}^{\infty} \frac{x}{x^{2}+1} dx \] Let's calculate each integral separately. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{1}^{+\infty} \frac{1}{x+1} dx\) - step1: Rewrite the improper integral: \(\lim _{a\rightarrow +\infty}\left(\int_{1}^{a} \frac{1}{x+1} dx\right)\) - step2: Evaluate the integral: \(\lim _{a\rightarrow +\infty}\left(\ln{\left(\frac{\left|a+1\right|}{2}\right)}\right)\) - step3: Rewrite the expression: \(\ln{\left(\lim _{a\rightarrow +\infty}\left(\frac{\left|a+1\right|}{2}\right)\right)}\) - step4: Calculate: \(\ln{\left(+\infty\right)}\) - step5: Calculate: \(+\infty\) - step6: Calculate: \(\textrm{Diverges}\) The first integral \( \int_{1}^{\infty} \frac{1}{x+1} dx \) diverges. Now, let's calculate the second integral \( \int_{1}^{\infty} \frac{x}{x^{2}+1} dx \). Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{1}^{+\infty} \frac{x}{x^{2}+1} dx\) - step1: Rewrite the improper integral: \(\lim _{a\rightarrow +\infty}\left(\int_{1}^{a} \frac{x}{x^{2}+1} dx\right)\) - step2: Evaluate the integral: \(\lim _{a\rightarrow +\infty}\left(\frac{1}{2}\ln{\left(a^{2}+1\right)}-\frac{\ln{\left(2\right)}}{2}\right)\) - step3: Rewrite the expression: \(\lim _{a\rightarrow +\infty}\left(\frac{1}{2}\ln{\left(a^{2}+1\right)}\right)+\lim _{a\rightarrow +\infty}\left(-\frac{\ln{\left(2\right)}}{2}\right)\) - step4: Calculate: \(\left(+\infty\right)-\frac{\ln{\left(2\right)}}{2}\) - step5: Calculate: \(+\infty\) - step6: Calculate: \(\textrm{Diverges}\)