Answer
$$\frac{1}{4}\ln 2 + \frac{1}{8}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{dy}}{{\left( {y + 1} \right)\left( {{y^2} + 1} \right)}}} \cr
& {\text{Decompose }}\frac{1}{{\left( {y + 1} \right)\left( {{y^2} + 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{1}{{\left( {y + 1} \right)\left( {{y^2} + 1} \right)}} = \frac{A}{{y + 1}} + \frac{{By + C}}{{{y^2} + 1}} \cr
& 1 = A\left( {{y^2} + 1} \right) + \left( {By + C} \right)\left( {y + 1} \right) \cr
& 1 = A{y^2} + A + B{y^2} + By + Cy + C \cr
& 1 = \left( {A{y^2} + B{y^2}} \right) + \left( {By + Cy} \right) + A + C \cr
& {\text{Equate the coefficients}} \cr
& A + B = 0,\,\,\,B + C = 0,\,\,\,\,A + C = 1 \cr
& {\text{Solving we obtain}} \cr
& A = \frac{1}{2},\,\,\,\,B = - \frac{1}{2},\,\,\,\,C = \frac{1}{2} \cr
& {\text{Therefore,}} \cr
& \frac{1}{{\left( {y + 1} \right)\left( {{y^2} + 1} \right)}} = \frac{1}{{2\left( {y + 1} \right)}} - \frac{{y - 1}}{{2\left( {{y^2} + 1} \right)}} \cr
& \int_0^1 {\frac{{dy}}{{\left( {y + 1} \right)\left( {{y^2} + 1} \right)}}} = \int_0^1 {\left[ {\frac{1}{{2\left( {y + 1} \right)}} - \frac{{y - 1}}{{2\left( {{y^2} + 1} \right)}}} \right]} dy \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int_0^1 {\frac{1}{{y + 1}}} dy - \frac{1}{2}\int_0^1 {\frac{y}{{{y^2} + 1}}} dy + \frac{1}{2}\int_0^1 {\frac{1}{{{y^2} + 1}}} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {\frac{1}{2}\ln \left| {y + 1} \right| - \frac{1}{4}\ln \left( {{y^2} + 1} \right) + \frac{1}{2}{{\tan }^{ - 1}}y} \right]_0^1 \cr
& {\text{ = }}\left[ {\frac{1}{2}\ln \left| {1 + 1} \right| - \frac{1}{4}\ln \left( {{1^2} + 1} \right) + \frac{1}{2}{{\tan }^{ - 1}}1} \right]\, - \left[ 0 \right] \cr
& = \frac{1}{2}\ln 2 - \frac{1}{4}\ln 2 + \frac{1}{2}\left( {\frac{\pi }{4}} \right) \cr
& = \frac{1}{4}\ln 2 + \frac{1}{8}\pi \cr} $$
