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人気のある微分積分 >

integral of \sqrt[3]{x^5}x^{-4/3}(x^3-1)

解

\int 3 x^{5} x^{- \frac{4}{3}} (x^{3} - 1) d x

解

\frac{3}{52} x^{\frac{4}{3}} (4 x^{3} - 13) + C

解答ステップ

\int 3 x^{5} x^{- \frac{4}{3}} (x^{3} - 1) d x

u置換積分法を適用する

= \int \frac{u ^{\frac{3}{5}} - 1}{5 u ^{\frac{11}{15}}} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int \frac{u ^{\frac{3}{5}} - 1}{u ^{\frac{11}{15}}} d u

拡張

\frac{u ^{\frac{3}{5}} - 1}{u ^{\frac{11}{15}}} : \frac{1}{u ^{\frac{2}{15}}} - \frac{1}{u ^{\frac{11}{15}}}

= \frac{1}{5} \cdot \int \frac{1}{u ^{\frac{2}{15}}} - \frac{1}{u ^{\frac{11}{15}}} d u

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

= \frac{1}{5} (\int \frac{1}{u ^{\frac{2}{15}}} d u - \int \frac{1}{u ^{\frac{11}{15}}} d u)

\int \frac{1}{u ^{\frac{2}{15}}} d u = \frac{15}{13} u^{\frac{13}{15}}

\int \frac{1}{u ^{\frac{11}{15}}} d u = \frac{15}{4} u^{\frac{4}{15}}

= \frac{1}{5} (\frac{15}{13} u^{\frac{13}{15}} - \frac{15}{4} u^{\frac{4}{15}})

代用を戻す

u = x^{5}

= \frac{1}{5} (\frac{15}{13} (x^{5})^{\frac{13}{15}} - \frac{15}{4} (x^{5})^{\frac{4}{15}})

簡素化

\frac{1}{5} (\frac{15}{13} (x^{5})^{\frac{13}{15}} - \frac{15}{4} (x^{5})^{\frac{4}{15}}) : \frac{3}{52} x^{\frac{4}{3}} (4 x^{3} - 13)

= \frac{3}{52} x^{\frac{4}{3}} (4 x^{3} - 13)

定数を解答に追加する

= \frac{3}{52} x^{\frac{4}{3}} (4 x^{3} - 13) + C

グラフ

人気の例

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