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受欢迎的微积分 >

integral of 1/((1-2x)^4sqrt(4x^2-4x-4))

解答

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

解答

- \frac{( 8 x ^{2} - 8 x + 7 ) x ^{2} - x - 1}{75 ( - 2 x + 1 ) ^{3}} + C

求解步骤

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

使用换元积分法

= \int - \frac{1}{2 u ^{4} u ^{2} - 5} 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{1}{u ^{4} u ^{2} - 5} d u

使用三角换元法

= - \frac{1}{2} \cdot \int \frac{1}{25 sec ^{3} ( v )} d v

提出常数:

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

= - \frac{1}{2} \cdot \frac{1}{25} \cdot \int \frac{1}{sec ^{3} ( v )} d v

使用三角恒等式改写

= - \frac{1}{2} \cdot \frac{1}{25} \cdot \int (1 - sin^{2} (v)) cos (v) d v

使用换元积分法

= - \frac{1}{2} \cdot \frac{1}{25} \cdot \int 1 - w^{2} d w

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} \cdot \frac{1}{25} (\int 1 d w - \int w^{2} d w)

\int 1 d w = w

\int w^{2} d w = \frac{w ^{3}}{3}

= - \frac{1}{2} \cdot \frac{1}{25} (w - \frac{w ^{3}}{3})

代回

= - \frac{1}{2} \cdot \frac{1}{25} sin (arcsec (\frac{1}{5} (1 - 2 x))) - \frac{sin ^{3} ( arcsec ( \frac{1}{5} ( 1 - 2 x ) ) )}{3}

化简

- \frac{1}{2} \cdot \frac{1}{25} sin (arcsec (\frac{1}{5} (1 - 2 x))) - \frac{sin ^{3} ( arcsec ( \frac{1}{5} ( 1 - 2 x ) ) )}{3} : - \frac{( 8 x ^{2} - 8 x + 7 ) x ^{2} - x - 1}{75 ( - 2 x + 1 ) ^{3}}

= - \frac{( 8 x ^{2} - 8 x + 7 ) x ^{2} - x - 1}{75 ( - 2 x + 1 ) ^{3}}

解答补常数

= - \frac{( 8 x ^{2} - 8 x + 7 ) x ^{2} - x - 1}{75 ( - 2 x + 1 ) ^{3}} + C

作图

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

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\int \frac{4}{65} x^{3} d x

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\int \frac{1}{x ^{2} + a x} d x

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\int \frac{( ln ( x ^{4} ) )}{x} d x