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受欢迎的微积分 >

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

解答

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

解答

\frac{( x ^{2} + 1 ) x ^{2} - 2}{6 x ^{3}} + C

求解步骤

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

使用三角换元法

= \int \frac{1}{4 sec ^{3} ( u )} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{1}{sec ^{3} ( u )} d u

使用三角恒等式改写

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

使用换元积分法

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

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

\int 1 d v = v

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

= \frac{1}{4} (v - \frac{v ^{3}}{3})

代回

= \frac{1}{4} sin (arcsec (\frac{1}{2} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{2} x ) )}{3}

化简

\frac{1}{4} sin (arcsec (\frac{1}{2} x)) - \frac{sin ^{3} ( arcsec ( \frac{1}{2} x ) )}{3} : \frac{( x ^{2} + 1 ) x ^{2} - 2}{6 x ^{3}}

= \frac{( x ^{2} + 1 ) x ^{2} - 2}{6 x ^{3}}

解答补常数

= \frac{( x ^{2} + 1 ) x ^{2} - 2}{6 x ^{3}} + C

作图

流行的例子

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\frac{d}{d x} (\frac{cos ( 3 x ^{3} )}{tan ( 3 x ^{2} + 2 x )})

tangent-3/2 x^2+4x+4

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\int x^{- \frac{1}{5}} d x

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\int \frac{12}{3 10 x + 4} - 3 d x

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