integral of (x^3)/(x^2-64)

解答

\int \frac{x ^{3}}{x ^{2} - 64} d x

\frac{1}{2} (x^{2} - 64 + 64 ln x^{2} - 64) + C

求解步骤

\int \frac{x ^{3}}{x ^{2} - 64} d x

使用换元积分法

= \int \frac{u + 64}{2 u} 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 + 64}{u} d u

乘开

\frac{u + 64}{u} : 1 + \frac{64}{u}

= \frac{1}{2} \cdot \int 1 + \frac{64}{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

= \frac{1}{2} (\int 1 d u + \int \frac{64}{u} d u)

\int 1 d u = u

\int \frac{64}{u} d u = 64 ln ∣ u ∣

= \frac{1}{2} (u + 64 ln ∣ u ∣)

u = x^{2} - 64

代回

= \frac{1}{2} (x^{2} - 64 + 64 ln x^{2} - 64)

解答补常数

= \frac{1}{2} (x^{2} - 64 + 64 ln x^{2} - 64) + C

t an g e n t f (x) = \frac{1}{x ^{2} + 1}, at x = 2

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