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受欢迎的微积分 >

integral of sqrt(4x^2-256)

解答

\int 4 x^{2} - 256 d x

解答

x x^{2} - 64 - 64 ln (\frac{1}{8} x + x^{2} - 64) + C

求解步骤

\int 4 x^{2} - 256 d x

使用三角换元法

= \int 128 tan^{2} (u) sec (u) d u

提出常数:

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

= 128 \cdot \int tan^{2} (u) sec (u) d u

使用三角恒等式改写

= 128 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

乘开

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

= 128 \cdot \int - sec (u) + sec^{3} (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

= 128 (- \int sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

= 128 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arcsec (\frac{2}{16} x)

代回

= 128 (- ln tan (arcsec (\frac{2}{16} x)) + sec (arcsec (\frac{2}{16} x)) + \frac{1}{2} sec (arcsec (\frac{2}{16} x)) tan (arcsec (\frac{2}{16} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{16} x)) + sec (arcsec (\frac{2}{16} x)))

化简

128 (- ln tan (arcsec (\frac{2}{16} x)) + sec (arcsec (\frac{2}{16} x)) + \frac{1}{2} sec (arcsec (\frac{2}{16} x)) tan (arcsec (\frac{2}{16} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{16} x)) + sec (arcsec (\frac{2}{16} x))) : x x^{2} - 64 - 64 ln (\frac{1}{8} x + x^{2} - 64)

= x x^{2} - 64 - 64 ln (\frac{1}{8} x + x^{2} - 64)

解答补常数

= x x^{2} - 64 - 64 ln (\frac{1}{8} x + x^{2} - 64) + C

作图

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