Answer
$$\frac{1}{4}\ln 2 + \frac{1}{8}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{dx}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}} \cr
& {\text{Decompose the integrand }}\frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& {\text{Multiply by }}\left( {x + 1} \right)\left( {{x^2} + 1} \right){\text{ and simplify}} \cr
& 1 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)\left( {x + 1} \right)\,\,\,\,\left( {\bf{1}} \right) \cr
& {\text{Substitute }}x = - 1{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& 1 = A\left( {{{\left( { - 1} \right)}^2} + 1} \right) + \left( {Bx + C} \right)\left( 0 \right) \cr
& A = \frac{1}{2} \cr
& {\text{Expand the equation }}\left( {\bf{1}} \right) \cr
& 1 = A{x^2} + A + B{x^2} + Bx + Cx + C \cr
& {\text{Group terms}} \cr
& 1 = \left( {A{x^2} + B{x^2}} \right) + \left( {Bx + Cx} \right) + A + C \cr
& {\text{Equating coefficients, we get}} \cr
& A + B = 0\,\,\,\, \to \,\,\,\,B = - \frac{1}{2},\,\,\,\,\,\,\,\,B + C = 0,\,\,\,C = \frac{1}{2} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{{1/2}}{{x + 1}} + \frac{{\left( { - 1/2} \right)x + 1/2}}{{{x^2} + 1}} \cr
& \frac{1}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \frac{{1/2}}{{x + 1}} - \frac{x}{{2\left( {{x^2} + 1} \right)}} + \frac{1}{{2\left( {{x^2} + 1} \right)}} \cr
& \int_0^1 {\frac{{dx}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}} = \int {\left( {\frac{{1/2}}{{x + 1}} - \frac{x}{{2\left( {{x^2} + 1} \right)}} + \frac{1}{{2\left( {{x^2} + 1} \right)}}} \right)} dx \cr
& {\text{Integrating, we get}} \cr
& = \left( {\frac{1}{2}\ln \left| {x + 1} \right| - \frac{1}{4}\ln \left( {{x^2} + 1} \right) + \frac{1}{2}\arctan x} \right)_0^1 \cr
& = \left( {\frac{1}{2}\ln \left| {1 + 1} \right| - \frac{1}{4}\ln \left( {{1^2} + 1} \right) + \frac{1}{2}\arctan 1} \right) - \left( {\frac{1}{2}\ln \left| {0 + 1} \right| - \frac{1}{4}\ln \left( {{0^2} + 1} \right) + \frac{1}{2}\arctan 0} \right) \cr
& = \frac{1}{2}\ln 2 - \frac{1}{4}\ln 2 + \frac{1}{2}\left( {\frac{\pi }{4}} \right) \cr
& = \frac{1}{4}\ln 2 + \frac{1}{8}\pi \cr} $$
