integral of t^{5/2}(t^2-4t+23)

解

\int t^{\frac{5}{2}} (t^{2} - 4 t + 23) d t

\frac{2}{11} t^{\frac{11}{2}} - \frac{8}{9} t^{\frac{9}{2}} + \frac{46}{7} t^{\frac{7}{2}} + C

解答ステップ

\int t^{\frac{5}{2}} (t^{2} - 4 t + 23) d t

拡張

t^{\frac{5}{2}} (t^{2} - 4 t + 23) : t^{\frac{9}{2}} - 4 t^{\frac{7}{2}} + 23 t^{\frac{5}{2}}

= \int t^{\frac{9}{2}} - 4 t^{\frac{7}{2}} + 23 t^{\frac{5}{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{9}{2}} d t - \int 4 t^{\frac{7}{2}} d t + \int 23 t^{\frac{5}{2}} d t

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

\int 4 t^{\frac{7}{2}} d t = \frac{8}{9} t^{\frac{9}{2}}

\int 23 t^{\frac{5}{2}} d t = \frac{46}{7} t^{\frac{7}{2}}

= \frac{2}{11} t^{\frac{11}{2}} - \frac{8}{9} t^{\frac{9}{2}} + \frac{46}{7} t^{\frac{7}{2}}

定数を解答に追加する

= \frac{2}{11} t^{\frac{11}{2}} - \frac{8}{9} t^{\frac{9}{2}} + \frac{46}{7} t^{\frac{7}{2}} + C