Answer
$$\ln \left( {\frac{{2\sqrt 2 }}{{\sqrt 5 }}} \right) + {\tan ^{ - 1}}2 - \frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{{x + 1}}{{x\left( {{x^2} + 1} \right)}}} dx \cr
& {\text{Integrating }}\int {\frac{{x + 1}}{{x\left( {{x^2} + 1} \right)}}} dx \cr
& \frac{{x + 1}}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& x + 1 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x \cr
& x + 1 = A{x^2} + A + B{x^2} + Cx \cr
& x + 1 = A{x^2} + B{x^2} + Cx + A \cr
& {\text{Comparing coefficients}} \cr
& \left( {A{x^2} + B{x^2}} \right) + Cx + A \cr
& {\text{Comparing coefficients}} \cr
& A + B = 0 \cr
& C = 1 \cr
& A = 1 \cr
& B = - 1 = - 1 \cr
& \frac{{x + 1}}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& \frac{{x + 1}}{{x\left( {{x^2} + 1} \right)}} = \frac{1}{x} + \frac{{ - x + 1}}{{{x^2} + 1}} \cr
& \int_1^2 {\frac{{x + 1}}{{x\left( {{x^2} + 1} \right)}}} dx = \int_1^2 {\left( {\frac{1}{x} + \frac{{ - x + 1}}{{{x^2} + 1}}} \right)} dx \cr
& = \int_1^2 {\left( {\frac{1}{x} - \frac{x}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right)} dx \cr
& = \left[ {\ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + {{\tan }^{ - 1}}x} \right]_1^2 \cr
& = \left[ {\ln \left| x \right| - \ln \sqrt {{x^2} + 1} + {{\tan }^{ - 1}}x} \right]_1^2 \cr
& = \left[ {\ln \frac{x}{{\sqrt {{x^2} + 1} }} + {{\tan }^{ - 1}}x} \right]_1^2 \cr
& = \left( {\ln \frac{2}{{\sqrt {{2^2} + 1} }} + {{\tan }^{ - 1}}2} \right) - \left( {\ln \frac{1}{{\sqrt {{1^2} + 1} }} + {{\tan }^{ - 1}}1} \right) \cr
& = \ln \frac{2}{{\sqrt 5 }} + {\tan ^{ - 1}}2 - \ln \frac{1}{{\sqrt 2 }} - \frac{\pi }{4} \cr
& = \ln \left( {\frac{2}{{\sqrt 5 }} \times \frac{{\sqrt 2 }}{1}} \right) + {\tan ^{ - 1}}2 - \frac{\pi }{4} \cr
& = \ln \left( {\frac{{2\sqrt 2 }}{{\sqrt 5 }}} \right) + {\tan ^{ - 1}}2 - \frac{\pi }{4} \approx 0.557 \cr} $$
