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受欢迎的微积分 >

integral of x^2(x^2+6)^{(-1)/2}

解答

\int x^{2} (x^{2} + 6)^{\frac{- 1}{2}} d x

解答

6 (- \frac{1}{2} ln \frac{1}{6 ^{\frac{1}{2}}} x + \frac{1}{6} (6 + x^{2}) + \frac{1}{12} x (6 + x^{2})^{\frac{1}{2}}) + C

求解步骤

\int x^{2} (x^{2} + 6)^{\frac{- 1}{2}} d x

使用三角换元法

= \int 6 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

= 6 \cdot \int tan^{2} (u) sec (u) d u

使用三角恒等式改写

= 6 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

乘开

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

= 6 \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

= 6 (- \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) ∣

= 6 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

u = arctan (\frac{1}{6 ^{\frac{1}{2}}} x)

代回

= 6 (- ln tan (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) + sec (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) + \frac{1}{2} sec (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) tan (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) + sec (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)))

化简

6 (- ln tan (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) + sec (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) + \frac{1}{2} sec (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) tan (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{6 ^{\frac{1}{2}}} x)) + sec (arctan (\frac{1}{6 ^{\frac{1}{2}}} x))) : 6 (- \frac{1}{2} ln \frac{1}{6 ^{\frac{1}{2}}} x + \frac{1}{6} (6 + x^{2}) + \frac{1}{12} x (6 + x^{2})^{\frac{1}{2}})

= 6 (- \frac{1}{2} ln \frac{1}{6 ^{\frac{1}{2}}} x + \frac{1}{6} (6 + x^{2}) + \frac{1}{12} x (6 + x^{2})^{\frac{1}{2}})

解答补常数

= 6 (- \frac{1}{2} ln \frac{1}{6 ^{\frac{1}{2}}} x + \frac{1}{6} (6 + x^{2}) + \frac{1}{12} x (6 + x^{2})^{\frac{1}{2}}) + C

作图

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