integral of sqrt(t)(t^4+t-4)

解

\int t (t^{4} + t - 4) d t

\frac{2}{11} t^{\frac{11}{2}} + \frac{2}{5} t^{\frac{5}{2}} - \frac{8}{3} t^{\frac{3}{2}} + C

解答ステップ

\int t (t^{4} + t - 4) d t

拡張

t (t^{4} + t - 4) : t^{\frac{9}{2}} + t^{\frac{3}{2}} - 4 t

= \int t^{\frac{9}{2}} + t^{\frac{3}{2}} - 4 t 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 t^{\frac{3}{2}} d t - \int 4 t d t

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

\int t^{\frac{3}{2}} d t = \frac{2}{5} t^{\frac{5}{2}}

\int 4 t d t = \frac{8}{3} t^{\frac{3}{2}}

= \frac{2}{11} t^{\frac{11}{2}} + \frac{2}{5} t^{\frac{5}{2}} - \frac{8}{3} t^{\frac{3}{2}}

定数を解答に追加する

= \frac{2}{11} t^{\frac{11}{2}} + \frac{2}{5} t^{\frac{5}{2}} - \frac{8}{3} t^{\frac{3}{2}} + C