Answer
$${\tan ^{ - 1}}y - \frac{1}{{{y^2} + 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{y^2} + 2y + 1}}{{{{\left( {{y^2} + 1} \right)}^2}}}} dy \cr
& {\text{Decompose the integrand }}\frac{{{y^2} + 2y + 1}}{{{{\left( {{y^2} + 1} \right)}^2}}}{\text{ into partial fractions}} \cr
& \frac{{{y^2} + 2y + 1}}{{{{\left( {{y^2} + 1} \right)}^2}}} = \frac{{Ay + B}}{{{y^2} + 1}} + \frac{{Cy + D}}{{{{\left( {{y^2} + 1} \right)}^2}}} \cr
& {\text{multiply by }}{\left( {{y^2} + 1} \right)^2}{\text{ and simplify}} \cr
& {y^2} + 2y + 1 = \left( {Ay + B} \right)\left( {{y^2} + 1} \right) + Cy + D \cr
& {\text{Expanding the equation }} \cr
& {y^2} + 2y + 1 = A{y^3} + Ay + B{y^2} + B + Cy + D \cr
& {\text{Group terms}} \cr
& {y^2} + 2y + 1 = A{y^3} + B{y^2} + Ay + Cy + B + D \cr
& {\text{Equating coefficients}} \cr
& A = 0,\,\,\,\,B = 1\,\,,\,\,\,\,A + C = 2,\,\,\,\,B + D = 1 \cr
& {\text{Solving, we obtain}} \cr
& {\text{ }}\,\,\,\,C = 2,\,\,\,D = 0 \cr
& \frac{{{y^2} + 2y + 1}}{{{{\left( {{y^2} + 1} \right)}^2}}} = \frac{1}{{{y^2} + 1}} + \frac{{2y}}{{{{\left( {{y^2} + 1} \right)}^2}}} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{{{y^2} + 2y + 1}}{{{{\left( {{y^2} + 1} \right)}^2}}}} dy = \int {\left( {\frac{1}{{{y^2} + 1}} + \frac{{2y}}{{{{\left( {{y^2} + 1} \right)}^2}}}} \right)} dy \cr
& = \int {\frac{1}{{{y^2} + 1}}} dy + \int {{{\left( {{y^2} + 1} \right)}^{ - 2}}\left( {2y} \right)} dy \cr
& {\text{Integrating gives}} \cr
& = {\tan ^{ - 1}}y - {\left( {{y^2} + 1} \right)^{ - 1}} + C \cr
& = {\tan ^{ - 1}}y - \frac{1}{{{y^2} + 1}} + C \cr} $$
