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受欢迎的微积分 >

integral of (x^2+8)/(xsqrt(x^2-4))

解答

\int \frac{x ^{2} + 8}{x x ^{2} - 4} d x

解答

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

求解步骤

\int \frac{x ^{2} + 8}{x x ^{2} - 4} d x

使用换元积分法

= \int \frac{u + 8}{2 u u - 4} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{u + 8}{u u - 4} d u

乘开

\frac{u + 8}{u u - 4} : \frac{1}{u - 4} + \frac{8}{u u - 4}

= \frac{1}{2} \cdot \int \frac{1}{u - 4} + \frac{8}{u u - 4} 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

= \frac{1}{2} (\int \frac{1}{u - 4} d u + \int \frac{8}{u u - 4} d u)

\int \frac{1}{u - 4} d u = 2 u - 4

\int \frac{8}{u u - 4} d u = 8 arctan (\frac{1}{2} u - 4)

= \frac{1}{2} (2 u - 4 + 8 arctan (\frac{1}{2} u - 4))

u = x^{2}

代回

= \frac{1}{2} (2 x^{2} - 4 + 8 arctan (\frac{1}{2} x^{2} - 4))

化简

\frac{1}{2} (2 x^{2} - 4 + 8 arctan (\frac{1}{2} x^{2} - 4)) : x^{2} - 4 + 4 arctan (\frac{1}{2} x^{2} - 4)

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

解答补常数

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

作图

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