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受欢迎的微积分 >

integral of 1/(x^5sqrt(4x^2-1))

解答

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

解答

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

求解步骤

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

使用三角换元法

= \int 16 cos^{4} (u) d u

提出常数:

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

= 16 \cdot \int cos^{4} (u) d u

化简

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

= 16 \cdot \int cos^{3} (u) cos (u) d u

使用分布积分法

= 16 (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))

= 16 (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

u = arcsec (2 x)

代回

= 16 (cos^{3} (arcsec (2 x)) sin (arcsec (2 x)) - (- \frac{3}{8} (arcsec (2 x) - \frac{1}{4} sin (4 arcsec (2 x)))))

化简

16 (cos^{3} (arcsec (2 x)) sin (arcsec (2 x)) - (- \frac{3}{8} (arcsec (2 x) - \frac{1}{4} sin (4 arcsec (2 x))))) : \frac{4 x ^{2} - 1}{x ^{4}} + 6 arcsec (2 x) - \frac{3}{2} sin (4 arcsec (2 x))

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

解答补常数

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

作图

流行的例子

integral of x/(sin^2(x))

\int \frac{x}{sin ^{2} ( x )} d x

derivative of arcsin(x)

\frac{d}{d x} (arcsin (x))

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(\partial)/(\partial x)((2xy^2)/(4+x^2y^2))

\frac{\partial}{\partial x} (\frac{2 x y ^{2}}{4 + x ^{2} y ^{2}})

limit as x approaches 0+of (e^{-1/x})/x

x \to 0 + lim (\frac{e ^{- \frac{1}{x}}}{x})