Answer
$$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{y - 1}}{2}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dy}}{{{y^2} - 2y + 5}}} \cr
& {\text{complete the square for }}{y^2} - 2y + 5 \cr
& = {y^2} - 2y + 1 + 4 \cr
& = {\left( {y - 1} \right)^2} + {2^2} \cr
& {\text{then}} \cr
& \int {\frac{{dy}}{{{y^2} - 2y + 5}}} = \int {\frac{{dy}}{{{{\left( {y - 1} \right)}^2} + {2^2}}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = y - 1,{\text{ so that }}du = dy \cr
& \int {\frac{{dy}}{{{{\left( {y - 1} \right)}^2} + {2^2}}}} = \int {\frac{{du}}{{{u^2} + {2^2}}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{u^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& {\text{write in terms of }}y;{\text{ replace }}y - 1{\text{ for }}u \cr
& = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{y - 1}}{2}} \right) + C \cr} $$
