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人気のある微分積分 >

integral of (1-x)/((1+x)\sqrt[3]{x)}

解

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

解

- \frac{3}{2} x^{\frac{2}{3}} - 2 ln 3 x + 1 + ln 4 x^{\frac{2}{3}} - 4 3 x + 4 + 23 arctan (\frac{1}{3} (2 3 x - 1)) + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{- u + 2}{u 3 u - 1} d u

拡張

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

= \int - \frac{1}{3 u - 1} + \frac{2}{u 3 u - 1} 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 \frac{1}{3 u - 1} d u + \int \frac{2}{u 3 u - 1} d u

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

\int \frac{2}{u 3 u - 1} d u = - 2 ln 3 u - 1 + 1 + ln 4 (u - 1)^{\frac{2}{3}} - 4 3 u - 1 + 4 + 23 arctan (\frac{1}{3} (2 3 u - 1 - 1))

= - \frac{3}{2} (u - 1)^{\frac{2}{3}} - 2 ln 3 u - 1 + 1 + ln 4 (u - 1)^{\frac{2}{3}} - 4 3 u - 1 + 4 + 23 arctan (\frac{1}{3} (2 3 u - 1 - 1))

代用を戻す

u = 1 + x

= - \frac{3}{2} (1 + x - 1)^{\frac{2}{3}} - 2 ln 3 1 + x - 1 + 1 + ln 4 (1 + x - 1)^{\frac{2}{3}} - 4 3 1 + x - 1 + 4 + 23 arctan (\frac{1}{3} (2 3 1 + x - 1 - 1))

簡素化

- \frac{3}{2} (1 + x - 1)^{\frac{2}{3}} - 2 ln 3 1 + x - 1 + 1 + ln 4 (1 + x - 1)^{\frac{2}{3}} - 4 3 1 + x - 1 + 4 + 23 arctan (\frac{1}{3} (2 3 1 + x - 1 - 1)) : - \frac{3}{2} x^{\frac{2}{3}} - 2 ln 3 x + 1 + ln 4 x^{\frac{2}{3}} - 4 3 x + 4 + 23 arctan (\frac{1}{3} (2 3 x - 1))

= - \frac{3}{2} x^{\frac{2}{3}} - 2 ln 3 x + 1 + ln 4 x^{\frac{2}{3}} - 4 3 x + 4 + 23 arctan (\frac{1}{3} (2 3 x - 1))

定数を解答に追加する

= - \frac{3}{2} x^{\frac{2}{3}} - 2 ln 3 x + 1 + ln 4 x^{\frac{2}{3}} - 4 3 x + 4 + 23 arctan (\frac{1}{3} (2 3 x - 1)) + C

グラフ

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