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

Calculate the integral \( \int_{1}^{\infty} \frac{x-1}{(x+1)(x^{2}+1)} dx \). Evaluate the integral by following steps: - step0: Evaluate: \(\int_{1}^{+\infty} \frac{x-1}{\left(x+1\right)\left(x^{2}+1\right)} dx\) - step1: Multiply the terms: \(\int_{1}^{+\infty} \frac{x-1}{x^{3}+x+x^{2}+1} dx\) - step2: Rewrite the improper integral: \(\lim _{a\rightarrow +\infty}\left(\int_{1}^{a} \frac{x-1}{x^{3}+x+x^{2}+1} dx\right)\) - step3: Evaluate the integral: \(\lim _{a\rightarrow +\infty}\left(\ln{\left(\frac{\left(a^{2}+1\right)^{\frac{1}{2}}}{\left|a+1\right|}\right)}+\frac{\ln{\left(2\right)}}{2}\right)\) - step4: Rewrite the expression: \(\lim _{a\rightarrow +\infty}\left(\ln{\left(\frac{\left(a^{2}+1\right)^{\frac{1}{2}}}{\left|a+1\right|}\right)}\right)+\lim _{a\rightarrow +\infty}\left(\frac{\ln{\left(2\right)}}{2}\right)\) - step5: Calculate: \(0+\frac{\ln{\left(2\right)}}{2}\) - step6: Calculate: \(\frac{\ln{\left(2\right)}}{2}\) The value of the integral \( I=\int_{1}^{\infty} \frac{x-1}{(x+1)(x^{2}+1)} dx \) is \( \frac{\ln{(2)}}{2} \). Therefore, the correct option is C. \( \ln \frac{2}{\sqrt{2}} \).