## Thomas' Calculus 13th Edition

$$\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{{y - 1}}{2}} \right) + C$$
\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}