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人気のある微分積分 >

integral of \sqrt[5]{x^4-x^2}(10x^3-5x)

解

\int 5 x^{4} - x^{2} (10 x^{3} - 5 x) d x

解

\frac{25}{12} (x^{4} - x^{2})^{\frac{6}{5}} + C

解答ステップ

\int 5 x^{4} - x^{2} (10 x^{3} - 5 x) d x

u置換積分法を適用する

= \int \frac{5 5 u}{2} d u

定数を除く:

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

= \frac{5}{2} \cdot \int 5 u d u

べき乗則を適用する

= \frac{5}{2} \cdot \frac{5}{6} u^{\frac{6}{5}}

代用を戻す

u = x^{4} - x^{2}

= \frac{5}{2} \cdot \frac{5}{6} (x^{4} - x^{2})^{\frac{6}{5}}

簡素化

\frac{5}{2} \cdot \frac{5}{6} (x^{4} - x^{2})^{\frac{6}{5}} : \frac{25}{12} (x^{4} - x^{2})^{\frac{6}{5}}

= \frac{25}{12} (x^{4} - x^{2})^{\frac{6}{5}}

定数を解答に追加する

= \frac{25}{12} (x^{4} - x^{2})^{\frac{6}{5}} + C

グラフ

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