integral of (x^2)/((6-x^2)^{3/2)}

解答

\int \frac{x ^{2}}{( 6 - x ^{2} ) ^{\frac{3}{2}}} d x

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

求解步骤

\int \frac{x ^{2}}{( 6 - x ^{2} ) ^{\frac{3}{2}}} d x

使用三角换元法

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

使用三角恒等式改写

= \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

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

\int 1 d u = u

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

= - u + tan (u)

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

代回

= - arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + tan (arcsin (\frac{1}{6 ^{\frac{1}{2}}} x))

化简

- arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + tan (arcsin (\frac{1}{6 ^{\frac{1}{2}}} x)) : - arcsin (\frac{1}{6 ^{\frac{1}{2}}} x) + \frac{x}{( 6 - x ^{2} ) ^{\frac{1}{2}}}

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

解答补常数

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

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