∫ 1 2 y + 5 y 7 y 3 d y = ∫ 1 2 ( y y 3 + 5 y 7 y 3 ) d y = ∫ 1 2 ( y − 2 + 5 y 4 ) d y = ( y − 2 + 1 − 2 + 1 + 5 y 4 + 1 4 + 1 ) | 1 2 = ( y − 1 − 1 + y 5 ) | 1 2 = ( − 1 y + y 5 ) | 1 2 = ( − 1 ( 2 ) + ( 2 ) 5 ) − ( − 1 ( 1 ) + ( 1 ) 5 ) = ( − 1 2 + 32 ) = ( − 1 2 + 64 2 ) = 63 2 {\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {y+5y^{7}}{y^{3}}}dy&=\int _{1}^{2}\left({\frac {y}{y^{3}}}+{\frac {5y^{7}}{y^{3}}}\right)dy=\int _{1}^{2}(y^{-2}+5y^{4})dy\\[2ex]&=\left({\frac {y^{-2+1}}{-2+1}}+{\frac {5y^{4+1}}{4+1}}\right){\bigg |}_{1}^{2}=\left({\frac {y^{-1}}{-1}}+y^{5}\right){\bigg |}_{1}^{2}=\left(-{\frac {1}{y}}+y^{5}\right){\bigg |}_{1}^{2}\\[2ex]&=\left(-{\frac {1}{(2)}}+(2)^{5}\right)-\left(-{\frac {1}{(1)}}+(1)^{5}\right)\\[2ex]&=\left(-{\frac {1}{2}}+32\right)=\left(-{\frac {1}{2}}+{\frac {64}{2}}\right)={\frac {63}{2}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{1}^{2}{\frac {y+5y^{7}}{y^{3}}}dy&=\int _{1}^{2}\left({\frac {y}{y^{3}}}+{\frac {5y^{7}}{y^{3}}}\right)dy=\int _{1}^{2}(y^{-2}+5y^{4})dy\\[2ex]&=\left({\frac {y^{-2+1}}{-2+1}}+{\frac {5y^{4+1}}{4+1}}\right){\bigg |}_{1}^{2}=\left({\frac {y^{-1}}{-1}}+y^{5}\right){\bigg |}_{1}^{2}=\left(-{\frac {1}{y}}+y^{5}\right){\bigg |}_{1}^{2}\\[2ex]&=\left(-{\frac {1}{(2)}}+(2)^{5}\right)-\left(-{\frac {1}{(1)}}+(1)^{5}\right)\\[2ex]&=\left(-{\frac {1}{2}}+32\right)=\left(-{\frac {1}{2}}+{\frac {64}{2}}\right)={\frac {63}{2}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cd78b59e19c4796743fcae9d4f7841d04dfd3f08)