integral of t^{-1/2}(t^4+2t^2-1)

解

\int t^{- \frac{1}{2}} (t^{4} + 2 t^{2} - 1) d t

\frac{2}{9} t^{\frac{9}{2}} + \frac{4}{5} t^{\frac{5}{2}} - 2 t^{\frac{1}{2}} + C

解答ステップ

\int t^{- \frac{1}{2}} (t^{4} + 2 t^{2} - 1) d t

簡素化

t^{- \frac{1}{2}} (t^{4} + 2 t^{2} - 1) : \frac{t ^{4} + 2 t ^{2} - 1}{t ^{\frac{1}{2}}}

= \int \frac{t ^{4} + 2 t ^{2} - 1}{t ^{\frac{1}{2}}} d t

拡張

\frac{t ^{4} + 2 t ^{2} - 1}{t ^{\frac{1}{2}}} : t^{\frac{7}{2}} + 2 t^{\frac{3}{2}} - \frac{1}{t ^{\frac{1}{2}}}

= \int t^{\frac{7}{2}} + 2 t^{\frac{3}{2}} - \frac{1}{t ^{\frac{1}{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{7}{2}} d t + \int 2 t^{\frac{3}{2}} d t - \int \frac{1}{t ^{\frac{1}{2}}} d t

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

\int 2 t^{\frac{3}{2}} d t = \frac{4}{5} t^{\frac{5}{2}}

\int \frac{1}{t ^{\frac{1}{2}}} d t = 2 t^{\frac{1}{2}}

= \frac{2}{9} t^{\frac{9}{2}} + \frac{4}{5} t^{\frac{5}{2}} - 2 t^{\frac{1}{2}}

定数を解答に追加する

= \frac{2}{9} t^{\frac{9}{2}} + \frac{4}{5} t^{\frac{5}{2}} - 2 t^{\frac{1}{2}} + C