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受欢迎的微积分 >

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

解答

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

解答

\frac{1}{16} (6 arcsin (2 x - 1) - sin (2 arcsin (2 x - 1)) - 16 x (- x + 1)) + C

求解步骤

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

对

x - x^{2}

配方:

- (x - \frac{1}{2})^{2} + \frac{1}{4}

= \int \frac{x ^{2}}{- ( x - \frac{1}{2} ) ^{2} + \frac{1}{4}} d x

使用换元积分法

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

提出常数:

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

= \frac{1}{2} \cdot \int \frac{( 2 u + 1 ) ^{2}}{- 4 u ^{2} + 1} d u

乘开

\frac{( 2 u + 1 ) ^{2}}{- 4 u ^{2} + 1} : \frac{4 u ^{2}}{- 4 u ^{2} + 1} + \frac{4 u}{- 4 u ^{2} + 1} + \frac{1}{- 4 u ^{2} + 1}

= \frac{1}{2} \cdot \int \frac{4 u ^{2}}{- 4 u ^{2} + 1} + \frac{4 u}{- 4 u ^{2} + 1} + \frac{1}{- 4 u ^{2} + 1} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{4 u ^{2}}{- 4 u ^{2} + 1} d u = \frac{1}{4} (arcsin (2 u) - \frac{1}{2} sin (2 arcsin (2 u)))

\int \frac{4 u}{- 4 u ^{2} + 1} d u = - - 4 u^{2} + 1

\int \frac{1}{- 4 u ^{2} + 1} d u = \frac{1}{2} arcsin (2 u)

= \frac{1}{2} (\frac{1}{4} (arcsin (2 u) - \frac{1}{2} sin (2 arcsin (2 u))) - - 4 u^{2} + 1 + \frac{1}{2} arcsin (2 u))

u = x - \frac{1}{2}

代回

= \frac{1}{2} \frac{1}{4} (arcsin (2 (x - \frac{1}{2})) - \frac{1}{2} sin (2 arcsin (2 (x - \frac{1}{2})))) - - 4 (x - \frac{1}{2})^{2} + 1 + \frac{1}{2} arcsin (2 (x - \frac{1}{2}))

化简

\frac{1}{2} \frac{1}{4} (arcsin (2 (x - \frac{1}{2})) - \frac{1}{2} sin (2 arcsin (2 (x - \frac{1}{2})))) - - 4 (x - \frac{1}{2})^{2} + 1 + \frac{1}{2} arcsin (2 (x - \frac{1}{2})) : \frac{1}{16} (6 arcsin (2 x - 1) - sin (2 arcsin (2 x - 1)) - 16 x (- x + 1))

= \frac{1}{16} (6 arcsin (2 x - 1) - sin (2 arcsin (2 x - 1)) - 16 x (- x + 1))

解答补常数

= \frac{1}{16} (6 arcsin (2 x - 1) - sin (2 arcsin (2 x - 1)) - 16 x (- x + 1)) + C

作图

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