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人気のある微分積分 >

integral of x^3sqrt(1+9x^2)

解

\int x^{3} 1 + 9 x^{2} d x

解

\frac{1}{1215} (27 x^{2} - 2) (9 x^{2} + 1)^{\frac{3}{2}} + C

解答ステップ

\int x^{3} 1 + 9 x^{2} d x

u置換積分法を適用する

= \int \frac{u ( u - 1 )}{162} d u

定数を除く:

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

= \frac{1}{162} \cdot \int u (u - 1) d u

拡張

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

= \frac{1}{162} \cdot \int 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}{162} (\int u^{\frac{3}{2}} d u - \int u d u)

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

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

= \frac{1}{162} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

代用を戻す

u = 1 + 9 x^{2}

= \frac{1}{162} (\frac{2}{5} (1 + 9 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 9 x^{2})^{\frac{3}{2}})

簡素化

\frac{1}{162} (\frac{2}{5} (1 + 9 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 9 x^{2})^{\frac{3}{2}}) : \frac{1}{1215} (27 x^{2} - 2) (9 x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{1215} (27 x^{2} - 2) (9 x^{2} + 1)^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{1215} (27 x^{2} - 2) (9 x^{2} + 1)^{\frac{3}{2}} + C

グラフ

人気の例

integral of 1+3x^2

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

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x \to 0 lim (\frac{e ^{x} + e ^{- x} - 2}{x sin ( x )})

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\int \frac{1}{x ^{2} + 8 x + 65} d x

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