integral of t^{7/2}(t^2-10t+17)

解

\int t^{\frac{7}{2}} (t^{2} - 10 t + 17) d t

\frac{2}{13} t^{\frac{13}{2}} - \frac{20}{11} t^{\frac{11}{2}} + \frac{34}{9} t^{\frac{9}{2}} + C

解答ステップ

\int t^{\frac{7}{2}} (t^{2} - 10 t + 17) d t

拡張

t^{\frac{7}{2}} (t^{2} - 10 t + 17) : t^{\frac{11}{2}} - 10 t^{\frac{9}{2}} + 17 t^{\frac{7}{2}}

= \int t^{\frac{11}{2}} - 10 t^{\frac{9}{2}} + 17 t^{\frac{7}{2}} d t

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t^{\frac{11}{2}} d t - \int 10 t^{\frac{9}{2}} d t + \int 17 t^{\frac{7}{2}} d t

\int t^{\frac{11}{2}} d t = \frac{2}{13} t^{\frac{13}{2}}

\int 10 t^{\frac{9}{2}} d t = \frac{20}{11} t^{\frac{11}{2}}

\int 17 t^{\frac{7}{2}} d t = \frac{34}{9} t^{\frac{9}{2}}

= \frac{2}{13} t^{\frac{13}{2}} - \frac{20}{11} t^{\frac{11}{2}} + \frac{34}{9} t^{\frac{9}{2}}

定数を解答に追加する

= \frac{2}{13} t^{\frac{13}{2}} - \frac{20}{11} t^{\frac{11}{2}} + \frac{34}{9} t^{\frac{9}{2}} + C