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人気のある微分積分 >

integral of x^3sqrt(2x+1)

解

\int x^{3} 2 x + 1 d x

解

\frac{1}{315} (35 x^{3} - 15 x^{2} + 6 x - 2) (2 x + 1)^{\frac{3}{2}} + C

解答ステップ

\int x^{3} 2 x + 1 d x

u置換積分法を適用する

= \int \frac{u ( u - 1 ) ^{3}}{16} d u

定数を除く:

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

= \frac{1}{16} \cdot \int u (u - 1)^{3} d u

拡張

u (u - 1)^{3} : u^{\frac{7}{2}} - 3 u^{\frac{5}{2}} + 3 u^{\frac{3}{2}} - u

= \frac{1}{16} \cdot \int u^{\frac{7}{2}} - 3 u^{\frac{5}{2}} + 3 u^{\frac{3}{2}} - u 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}{16} (\int u^{\frac{7}{2}} d u - \int 3 u^{\frac{5}{2}} d u + \int 3 u^{\frac{3}{2}} d u - \int u d u)

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

\int 3 u^{\frac{5}{2}} d u = \frac{6}{7} u^{\frac{7}{2}}

\int 3 u^{\frac{3}{2}} d u = \frac{6}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{16} (\frac{2}{9} u^{\frac{9}{2}} - \frac{6}{7} u^{\frac{7}{2}} + \frac{6}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

代用を戻す

u = 2 x + 1

= \frac{1}{16} (\frac{2}{9} (2 x + 1)^{\frac{9}{2}} - \frac{6}{7} (2 x + 1)^{\frac{7}{2}} + \frac{6}{5} (2 x + 1)^{\frac{5}{2}} - \frac{2}{3} (2 x + 1)^{\frac{3}{2}})

簡素化

\frac{1}{16} (\frac{2}{9} (2 x + 1)^{\frac{9}{2}} - \frac{6}{7} (2 x + 1)^{\frac{7}{2}} + \frac{6}{5} (2 x + 1)^{\frac{5}{2}} - \frac{2}{3} (2 x + 1)^{\frac{3}{2}}) : \frac{1}{315} (35 x^{3} - 15 x^{2} + 6 x - 2) (2 x + 1)^{\frac{3}{2}}

= \frac{1}{315} (35 x^{3} - 15 x^{2} + 6 x - 2) (2 x + 1)^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{315} (35 x^{3} - 15 x^{2} + 6 x - 2) (2 x + 1)^{\frac{3}{2}} + C

グラフ

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