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人気のある微分積分 >

integral of x(2x+5)^{10}

解

\int x (2 x + 5)^{10} d x

解

\frac{1}{528} (11 (2 x + 5)^{12} - 60 (2 x + 5)^{11}) + C

解答ステップ

\int x (2 x + 5)^{10} d x

u置換積分法を適用する

= \int \frac{u ^{10} ( 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^{10} (u - 5) d u

拡張

u^{10} (u - 5) : u^{11} - 5 u^{10}

= \frac{1}{4} \cdot \int u^{11} - 5 u^{10} 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^{11} d u - \int 5 u^{10} d u)

\int u^{11} d u = \frac{u ^{12}}{12}

\int 5 u^{10} d u = \frac{5 u ^{11}}{11}

= \frac{1}{4} (\frac{u ^{12}}{12} - \frac{5 u ^{11}}{11})

代用を戻す

u = 2 x + 5

= \frac{1}{4} (\frac{( 2 x + 5 ) ^{12}}{12} - \frac{5 ( 2 x + 5 ) ^{11}}{11})

簡素化

\frac{1}{4} (\frac{( 2 x + 5 ) ^{12}}{12} - \frac{5 ( 2 x + 5 ) ^{11}}{11}) : \frac{1}{528} (11 (2 x + 5)^{12} - 60 (2 x + 5)^{11})

= \frac{1}{528} (11 (2 x + 5)^{12} - 60 (2 x + 5)^{11})

定数を解答に追加する

= \frac{1}{528} (11 (2 x + 5)^{12} - 60 (2 x + 5)^{11}) + C

グラフ

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