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受欢迎的微积分 >

integral of x^2sqrt(x^2-2)

解答

\int x^{2} x^{2} - 2 d x

解答

- \frac{1}{4} x x^{2} - 2 - \frac{1}{2} ln \frac{1}{2} (x^{2} - 2) + \frac{1}{2} x + \frac{1}{4} x^{3} x^{2} - 2 + C

求解步骤

\int x^{2} x^{2} - 2 d x

使用三角换元法

= \int 4 sec^{3} (u) tan^{2} (u) d u

提出常数:

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

= 4 \cdot \int sec^{3} (u) tan^{2} (u) d u

使用三角恒等式改写

= 4 \cdot \int sec^{3} (u) (- 1 + sec^{2} (u)) d u

乘开

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= 4 \cdot \int - sec^{3} (u) + sec^{5} (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

= 4 (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

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

\int sec^{5} (u) d u = \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) ∣)

= 4 (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \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}{2} x)

代回

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

化简

4 - (\frac{1}{2} sec (arcsec (\frac{1}{2} x)) tan (arcsec (\frac{1}{2} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))) + \frac{sec ^{4} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arcsec (\frac{1}{2} x)) tan (arcsec (\frac{1}{2} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))) : - \frac{1}{4} x x^{2} - 2 - \frac{1}{2} ln \frac{1}{2} (x^{2} - 2) + \frac{1}{2} x + \frac{1}{4} x^{3} x^{2} - 2

= - \frac{1}{4} x x^{2} - 2 - \frac{1}{2} ln \frac{1}{2} (x^{2} - 2) + \frac{1}{2} x + \frac{1}{4} x^{3} x^{2} - 2

解答补常数

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

作图

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