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受欢迎的微积分 >

integral of (y^4)/(sqrt(y^2-9))

解答

\int \frac{y ^{4}}{y ^{2} - 9} d y

解答

81 \frac{1}{324} y^{3} y^{2} - 9 + \frac{1}{24} y y^{2} - 9 + 9 ln \frac{y + y ^{2} - 9}{3} + C

求解步骤

\int \frac{y ^{4}}{y ^{2} - 9} d y

化简

y^{4} : y^{3} y

= \int \frac{y ^{3} y}{y ^{2} - 9} d y

使用三角换元法

= \int 81 sec^{5} (u) d u

提出常数:

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

= 81 \cdot \int sec^{5} (u) d u

使用降次积分法

= 81 (\frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} \cdot \int sec^{3} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 81 (\frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

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

代回

= 81 (\frac{sec ^{4} ( arcsec ( \frac{1}{3} y ) ) sin ( arcsec ( \frac{1}{3} y ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arcsec (\frac{1}{3} y)) tan (arcsec (\frac{1}{3} y)) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} y)) + sec (arcsec (\frac{1}{3} y))))

化简

81 (\frac{sec ^{4} ( arcsec ( \frac{1}{3} y ) ) sin ( arcsec ( \frac{1}{3} y ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arcsec (\frac{1}{3} y)) tan (arcsec (\frac{1}{3} y)) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} y)) + sec (arcsec (\frac{1}{3} y)))) : 81 \frac{1}{324} y^{3} y^{2} - 9 + \frac{1}{24} y y^{2} - 9 + 9 ln \frac{y + y ^{2} - 9}{3}

= 81 \frac{1}{324} y^{3} y^{2} - 9 + \frac{1}{24} y y^{2} - 9 + 9 ln \frac{y + y ^{2} - 9}{3}

解答补常数

= 81 \frac{1}{324} y^{3} y^{2} - 9 + \frac{1}{24} y y^{2} - 9 + 9 ln \frac{y + y ^{2} - 9}{3} + C

作图

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