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人気のある微分積分 >

integral of x(5x+1)^{19}

解

\int x (5 x + 1)^{19} d x

解

\frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20}) + C

解答ステップ

\int x (5 x + 1)^{19} d x

u置換積分法を適用する

= \int \frac{u ^{19} ( u - 1 )}{25} d u

定数を除く:

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

= \frac{1}{25} \cdot \int u^{19} (u - 1) d u

拡張

u^{19} (u - 1) : u^{20} - u^{19}

= \frac{1}{25} \cdot \int u^{20} - u^{19} 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}{25} (\int u^{20} d u - \int u^{19} d u)

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

\int u^{19} d u = \frac{u ^{20}}{20}

= \frac{1}{25} (\frac{u ^{21}}{21} - \frac{u ^{20}}{20})

代用を戻す

u = 5 x + 1

= \frac{1}{25} (\frac{( 5 x + 1 ) ^{21}}{21} - \frac{( 5 x + 1 ) ^{20}}{20})

簡素化

\frac{1}{25} (\frac{( 5 x + 1 ) ^{21}}{21} - \frac{( 5 x + 1 ) ^{20}}{20}) : \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20})

= \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20})

定数を解答に追加する

= \frac{1}{10500} (20 (5 x + 1)^{21} - 21 (5 x + 1)^{20}) + C

グラフ

人気の例

integral of e^{-bt}

\int e^{- b t} d t

integral of 5/(9+5x)

\int \frac{5}{9 + 5 x} d x

integral from 0 to a of x^2e^x

\int_{0}^{a} x^{2} e^{x} d x

(\partial)/(\partial x)(6x^3y^2)

\frac{\partial}{\partial x} (6 x^{3} y^{2})

limit as x approaches 1 of (5x-1)^{-2}

x \to 1 lim ((5 x - 1)^{- 2})