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integral of 1/(x^5sqrt(x^2-9))

解答

\int \frac{1}{x ^{5} x ^{2} - 9} d x

解答

\frac{1}{243} (\frac{27 x ^{2} - 9}{x ^{4}} + \frac{3}{8} (arcsec (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsec (\frac{1}{3} x)))) + C

求解步骤

\int \frac{1}{x ^{5} x ^{2} - 9} d x

使用三角换元法

= \int \frac{1}{243 sec ^{4} ( u )} d u

提出常数:

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

= \frac{1}{243} \cdot \int \frac{1}{sec ^{4} ( u )} d u

使用三角恒等式改写

= \frac{1}{243} \cdot \int cos^{4} (u) d u

化简

cos^{4} (u) : cos^{3} (u) cos (u)

= \frac{1}{243} \cdot \int cos^{3} (u) cos (u) d u

使用分布积分法

= \frac{1}{243} (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= \frac{1}{243} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

u = arcsec (\frac{1}{3} x)

代回

= \frac{1}{243} (cos^{3} (arcsec (\frac{1}{3} x)) sin (arcsec (\frac{1}{3} x)) - (- \frac{3}{8} (arcsec (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsec (\frac{1}{3} x)))))

化简

\frac{1}{243} (cos^{3} (arcsec (\frac{1}{3} x)) sin (arcsec (\frac{1}{3} x)) - (- \frac{3}{8} (arcsec (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsec (\frac{1}{3} x))))) : \frac{1}{243} (\frac{27 x ^{2} - 9}{x ^{4}} + \frac{3}{8} (arcsec (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsec (\frac{1}{3} x))))

= \frac{1}{243} (\frac{27 x ^{2} - 9}{x ^{4}} + \frac{3}{8} (arcsec (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsec (\frac{1}{3} x))))

解答补常数

= \frac{1}{243} (\frac{27 x ^{2} - 9}{x ^{4}} + \frac{3}{8} (arcsec (\frac{1}{3} x) - \frac{1}{4} sin (4 arcsec (\frac{1}{3} x)))) + C

作图

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