integral of 3/(sqrt(x^2-9))

解答

\int \frac{3}{x ^{2} - 9} d x

3 ln (\frac{1}{3} x^{2} - 9 + x) + C

求解步骤

\int \frac{3}{x ^{2} - 9} d x

提出常数:

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

= 3 \cdot \int \frac{1}{x ^{2} - 9} d x

使用三角换元法

= 3 \cdot \int sec (u) d u

使用积分常见定则:

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

= 3 ln ∣ tan (u) + sec (u) ∣

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

代回

= 3 ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x))

化简

3 ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x)) : 3 ln (\frac{1}{3} x^{2} - 9 + x)

= 3 ln (\frac{1}{3} x^{2} - 9 + x)

解答补常数

= 3 ln (\frac{1}{3} x^{2} - 9 + x) + C

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