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人気のある微分積分 >

integral of 1/(x^3*\sqrt[5]{1+1/x)}

解

\int \frac{1}{x ^{3} \cdot 5 1 + \frac{1}{x}} d x

解

- \frac{5}{9} (\frac{x + 1}{x})^{\frac{9}{5}} + \frac{5}{4} (\frac{x + 1}{x})^{\frac{4}{5}} + C

解答ステップ

\int \frac{1}{x ^{3} 5 1 + \frac{1}{x}} d x

u置換積分法を適用する

= \int - \frac{u - 1}{5 u} d u

定数を除く:

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

= - \int \frac{u - 1}{5 u} d u

拡張

\frac{u - 1}{5 u} : u^{\frac{4}{5}} - \frac{1}{5 u}

= - \int u^{\frac{4}{5}} - \frac{1}{5 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

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

\int u^{\frac{4}{5}} d u = \frac{5}{9} u^{\frac{9}{5}}

\int \frac{1}{5 u} d u = \frac{5}{4} u^{\frac{4}{5}}

= - (\frac{5}{9} u^{\frac{9}{5}} - \frac{5}{4} u^{\frac{4}{5}})

代用を戻す

u = 1 + \frac{1}{x}

= - (\frac{5}{9} (1 + \frac{1}{x})^{\frac{9}{5}} - \frac{5}{4} (1 + \frac{1}{x})^{\frac{4}{5}})

簡素化

- (\frac{5}{9} (1 + \frac{1}{x})^{\frac{9}{5}} - \frac{5}{4} (1 + \frac{1}{x})^{\frac{4}{5}}) : - \frac{5}{9} (\frac{x + 1}{x})^{\frac{9}{5}} + \frac{5}{4} (\frac{x + 1}{x})^{\frac{4}{5}}

= - \frac{5}{9} (\frac{x + 1}{x})^{\frac{9}{5}} + \frac{5}{4} (\frac{x + 1}{x})^{\frac{4}{5}}

定数を解答に追加する

= - \frac{5}{9} (\frac{x + 1}{x})^{\frac{9}{5}} + \frac{5}{4} (\frac{x + 1}{x})^{\frac{4}{5}} + C

グラフ

人気の例

tangent f(x)=-2x^{2/3}-4,\at x=8

t an g e n t f (x) = - 2 x^{\frac{2}{3}} - 4, at x = 8

integral of (x^3)/(1-x^8)

\int \frac{x ^{3}}{1 - x ^{8}} d x

derivative (1-2ln(x))/(x^3)

d er i v a t i v e \frac{1 - 2 ln ( x )}{x ^{3}}

integral of (sqrt(64-x^2))/(x^2)

\int \frac{64 - x ^{2}}{x ^{2}} d x

integral from 1 to 3 of 9/(x^2)

\int_{1}^{3} \frac{9}{x ^{2}} d x