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人気のある微分積分 >

integral of x(2x+5)^{20}

解

\int x (2 x + 5)^{20} d x

解

\frac{1}{1848} (21 (2 x + 5)^{22} - 110 (2 x + 5)^{21}) + C

解答ステップ

\int x (2 x + 5)^{20} d x

u置換積分法を適用する

= \int \frac{u ^{20} ( u - 5 )}{4} d u

定数を除く:

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

= \frac{1}{4} \cdot \int u^{20} (u - 5) d u

拡張

u^{20} (u - 5) : u^{21} - 5 u^{20}

= \frac{1}{4} \cdot \int u^{21} - 5 u^{20} 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}{4} (\int u^{21} d u - \int 5 u^{20} d u)

\int u^{21} d u = \frac{u ^{22}}{22}

\int 5 u^{20} d u = \frac{5 u ^{21}}{21}

= \frac{1}{4} (\frac{u ^{22}}{22} - \frac{5 u ^{21}}{21})

代用を戻す

u = 2 x + 5

= \frac{1}{4} (\frac{( 2 x + 5 ) ^{22}}{22} - \frac{5 ( 2 x + 5 ) ^{21}}{21})

簡素化

\frac{1}{4} (\frac{( 2 x + 5 ) ^{22}}{22} - \frac{5 ( 2 x + 5 ) ^{21}}{21}) : \frac{1}{1848} (21 (2 x + 5)^{22} - 110 (2 x + 5)^{21})

= \frac{1}{1848} (21 (2 x + 5)^{22} - 110 (2 x + 5)^{21})

定数を解答に追加する

= \frac{1}{1848} (21 (2 x + 5)^{22} - 110 (2 x + 5)^{21}) + C

グラフ

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