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人気のある微分積分 >

integral of 1/(sqrt(x)(1-\sqrt{1-x))}

解

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

解

2 (arcsin (- - x + 1) + \frac{2}{tan ( \frac{1}{2} arcsin ( - - x + 1 ) ) + 1}) + C

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

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

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

拡張

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

= 2 \cdot \int \frac{1}{- u ^{2} + 2 u} - \frac{1}{u - u ^{2} + 2 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

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

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

\int \frac{1}{u - u ^{2} + 2 u} d u = - \frac{2}{tan ( \frac{1}{2} arcsin ( u - 1 ) ) + 1}

= 2 (arcsin (u - 1) - (- \frac{2}{tan ( \frac{1}{2} arcsin ( u - 1 ) ) + 1}))

代用を戻す

u = 1 - 1 - x

= 2 (arcsin (1 - 1 - x - 1) - (- \frac{2}{tan ( \frac{1}{2} arcsin ( 1 - 1 - x - 1 ) ) + 1}))

簡素化

2 (arcsin (1 - 1 - x - 1) - (- \frac{2}{tan ( \frac{1}{2} arcsin ( 1 - 1 - x - 1 ) ) + 1})) : 2 (arcsin (- - x + 1) + \frac{2}{tan ( \frac{1}{2} arcsin ( - - x + 1 ) ) + 1})

= 2 (arcsin (- - x + 1) + \frac{2}{tan ( \frac{1}{2} arcsin ( - - x + 1 ) ) + 1})

定数を解答に追加する

= 2 (arcsin (- - x + 1) + \frac{2}{tan ( \frac{1}{2} arcsin ( - - x + 1 ) ) + 1}) + C

グラフ

人気の例

integral of (8-3x)/((x+1)(x^2-4x+6))

\int \frac{8 - 3 x}{( x + 1 ) ( x ^{2} - 4 x + 6 )} d x

integral of (x^2-x+6)/((x)(x^2+3))

\int \frac{x ^{2} - x + 6}{( x ) ( x ^{2} + 3 )} d x

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\int 2 cos^{3} (x) sin (x) d x

integral of (2x)/(sqrt(x^4+9))

\int \frac{2 x}{x ^{4} + 9} d x

integral of (sin(4e^{-x}))/(e^x)

\int \frac{sin ( 4 e ^{- x} )}{e ^{x}} d x