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人気のある微分積分 >

integral of x^5\sqrt[3]{(1+x^3)^2}

解

\int x^{5} 3 (1 + x^{3})^{2} d x

解

\frac{1}{40} (8 x^{3} (x^{3} + 1)^{\frac{5}{3}} - 3 (x^{3} + 1)^{\frac{8}{3}}) + C

解答ステップ

\int x^{5} 3 (1 + x^{3})^{2} d x

3 (1 + x^{3})^{2} = (1 + x^{3})^{\frac{2}{3}}

= \int x^{5} (1 + x^{3})^{\frac{2}{3}} d x

u置換積分法を適用する

= \int \frac{u ^{\frac{1}{5}} ( u ^{\frac{3}{5}} + 1 ) ^{\frac{2}{3}}}{5} 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 u^{\frac{1}{5}} (u^{\frac{3}{5}} + 1)^{\frac{2}{3}} d u

u置換積分法を適用する

= \frac{1}{5} \cdot \int \frac{5 ( v - 1 ) v ^{\frac{2}{3}}}{3} d v

定数を除く:

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

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int (v - 1) v^{\frac{2}{3}} d v

部分積分を適用する

= \frac{1}{5} \cdot \frac{5}{3} (\frac{3}{5} v^{\frac{5}{3}} (v - 1) - \int \frac{3}{5} v^{\frac{5}{3}} d v)

\int \frac{3}{5} v^{\frac{5}{3}} d v = \frac{9}{40} v^{\frac{8}{3}}

= \frac{1}{5} \cdot \frac{5}{3} (\frac{3}{5} v^{\frac{5}{3}} (v - 1) - \frac{9}{40} v^{\frac{8}{3}})

代用を戻す

= \frac{1}{5} \cdot \frac{5}{3} (\frac{3}{5} ((x^{5})^{\frac{3}{5}} + 1)^{\frac{5}{3}} ((x^{5})^{\frac{3}{5}} + 1 - 1) - \frac{9}{40} ((x^{5})^{\frac{3}{5}} + 1)^{\frac{8}{3}})

簡素化

\frac{1}{5} \cdot \frac{5}{3} (\frac{3}{5} ((x^{5})^{\frac{3}{5}} + 1)^{\frac{5}{3}} ((x^{5})^{\frac{3}{5}} + 1 - 1) - \frac{9}{40} ((x^{5})^{\frac{3}{5}} + 1)^{\frac{8}{3}}) : \frac{1}{40} (8 x^{3} (x^{3} + 1)^{\frac{5}{3}} - 3 (x^{3} + 1)^{\frac{8}{3}})

= \frac{1}{40} (8 x^{3} (x^{3} + 1)^{\frac{5}{3}} - 3 (x^{3} + 1)^{\frac{8}{3}})

定数を解答に追加する

= \frac{1}{40} (8 x^{3} (x^{3} + 1)^{\frac{5}{3}} - 3 (x^{3} + 1)^{\frac{8}{3}}) + C

グラフ

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