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受欢迎的微积分 >

integral of 1/(x^3sqrt(9x^2-1))

解答

\int \frac{1}{x ^{3} 9 x ^{2} - 1} d x

解答

\frac{9}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x))) + C

求解步骤

\int \frac{1}{x ^{3} 9 x ^{2} - 1} d x

使用三角换元法

= \int 9 cos^{2} (u) d u

提出常数:

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

= 9 \cdot \int cos^{2} (u) d u

使用三角恒等式改写

= 9 \cdot \int \frac{1 + cos ( 2 u )}{2} d u

提出常数:

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

= 9 \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) 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

= 9 \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 9 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

u = arcsec (3 x)

代回

= 9 \cdot \frac{1}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x)))

化简

9 \cdot \frac{1}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x))) : \frac{9}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x)))

= \frac{9}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x)))

解答补常数

= \frac{9}{2} (arcsec (3 x) + \frac{1}{2} sin (2 arcsec (3 x))) + C

作图

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