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人気のある微分積分 >

integral of x/((3x+1)^{1/3)}

解

\int \frac{x}{( 3 x + 1 ) ^{\frac{1}{3}}} d x

解

\frac{1}{10} (2 x - 1) (3 x + 1)^{\frac{2}{3}} + C

解答ステップ

\int \frac{x}{( 3 x + 1 ) ^{\frac{1}{3}}} d x

指数の規則を適用する:

\frac{1}{a ^{b}} = a^{- b}

\frac{x}{( 3 x + 1 ) ^{\frac{1}{3}}} = (x) (3 x + 1)^{- \frac{1}{3}}

= \int x (3 x + 1)^{- \frac{1}{3}} d x

u置換積分法を適用する

= \int \frac{u - 1}{9 u ^{\frac{1}{3}}} d u

定数を除く:

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

= \frac{1}{9} \cdot \int \frac{u - 1}{u ^{\frac{1}{3}}} d u

拡張

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

= \frac{1}{9} \cdot \int u^{\frac{2}{3}} - \frac{1}{u ^{\frac{1}{3}}} 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}{9} (\int u^{\frac{2}{3}} d u - \int \frac{1}{u ^{\frac{1}{3}}} d u)

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

\int \frac{1}{u ^{\frac{1}{3}}} d u = \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{9} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

代用を戻す

u = 3 x + 1

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

簡素化

\frac{1}{9} (\frac{3}{5} (3 x + 1)^{\frac{5}{3}} - \frac{3}{2} (3 x + 1)^{\frac{2}{3}}) : \frac{1}{10} (2 x - 1) (3 x + 1)^{\frac{2}{3}}

= \frac{1}{10} (2 x - 1) (3 x + 1)^{\frac{2}{3}}

定数を解答に追加する

= \frac{1}{10} (2 x - 1) (3 x + 1)^{\frac{2}{3}} + C

グラフ

人気の例

integral of u/(u^2+9)

\int \frac{u}{u ^{2} + 9} d u

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\int (\frac{( x ^{3} - 1 )}{x})^{2} d x

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\int \frac{1}{2} e^{\frac{x}{2}} d x

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