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受欢迎的微积分 >

integral of (sqrt(4x^2-16))/x

解答

\int \frac{4 x ^{2} - 16}{x} d x

解答

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

求解步骤

\int \frac{4 x ^{2} - 16}{x} d x

使用三角换元法

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

使用三角恒等式改写

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

\int 1 d u = u

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

= 4 (- u + tan (u))

u = arcsec (\frac{2}{4} x)

代回

= 4 (- arcsec (\frac{2}{4} x) + tan (arcsec (\frac{2}{4} x)))

化简

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

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

解答补常数

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

作图

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