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受欢迎的微积分 >

integral of 1/(x^4sqrt(4x^2-9))

解答

\int \frac{1}{x ^{4} 4 x ^{2} - 9} d x

解答

\frac{8}{81} \frac{4 x ^{2} - 9}{2 x} - \frac{( 4 x ^{2} - 9 ) ^{\frac{3}{2}}}{24 x ^{3}} + C

求解步骤

\int \frac{1}{x ^{4} 4 x ^{2} - 9} d x

使用三角换元法

= \int \frac{8}{81 sec ^{3} ( u )} d u

提出常数:

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

= \frac{8}{81} \cdot \int \frac{1}{sec ^{3} ( u )} d u

使用三角恒等式改写

= \frac{8}{81} \cdot \int (1 - sin^{2} (u)) cos (u) d u

使用换元积分法

= \frac{8}{81} \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{8}{81} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

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

= \frac{8}{81} (v - \frac{v ^{3}}{3})

代回

= \frac{8}{81} (sin (arcsec (\frac{2}{3} x)) - \frac{sin ^{3} ( arcsec ( \frac{2}{3} x ) )}{3})

化简

\frac{8}{81} (sin (arcsec (\frac{2}{3} x)) - \frac{sin ^{3} ( arcsec ( \frac{2}{3} x ) )}{3}) : \frac{8}{81} \frac{4 x ^{2} - 9}{2 x} - \frac{( 4 x ^{2} - 9 ) ^{\frac{3}{2}}}{24 x ^{3}}

= \frac{8}{81} \frac{4 x ^{2} - 9}{2 x} - \frac{( 4 x ^{2} - 9 ) ^{\frac{3}{2}}}{24 x ^{3}}

解答补常数

= \frac{8}{81} \frac{4 x ^{2} - 9}{2 x} - \frac{( 4 x ^{2} - 9 ) ^{\frac{3}{2}}}{24 x ^{3}} + C

作图

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