integral of (2x^5-x^2)^2(x^{-1/2})

解

\int (2 x^{5} - x^{2})^{2} (x^{- \frac{1}{2}}) d x

\frac{8}{21} x^{\frac{21}{2}} - \frac{8}{15} x^{\frac{15}{2}} + \frac{2}{9} x^{\frac{9}{2}} + C

解答ステップ

\int (2 x^{5} - x^{2})^{2} x^{- \frac{1}{2}} d x

拡張

(2 x^{5} - x^{2})^{2} x^{- \frac{1}{2}} : 4 x^{\frac{19}{2}} - 4 x^{\frac{13}{2}} + x^{\frac{7}{2}}

= \int 4 x^{\frac{19}{2}} - 4 x^{\frac{13}{2}} + x^{\frac{7}{2}} d x

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 4 x^{\frac{19}{2}} d x - \int 4 x^{\frac{13}{2}} d x + \int x^{\frac{7}{2}} d x

\int 4 x^{\frac{19}{2}} d x = \frac{8}{21} x^{\frac{21}{2}}

\int 4 x^{\frac{13}{2}} d x = \frac{8}{15} x^{\frac{15}{2}}

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

= \frac{8}{21} x^{\frac{21}{2}} - \frac{8}{15} x^{\frac{15}{2}} + \frac{2}{9} x^{\frac{9}{2}}

定数を解答に追加する

= \frac{8}{21} x^{\frac{21}{2}} - \frac{8}{15} x^{\frac{15}{2}} + \frac{2}{9} x^{\frac{9}{2}} + C