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受欢迎的微积分 >

integral of 1/(5000x-x^2)

解答

\int \frac{1}{5000 x - x ^{2}} d x

解答

\frac{1}{5000} (ln \frac{x - 2500}{2500} + 1 - ln \frac{x - 2500}{2500} - 1) + C

求解步骤

\int \frac{1}{5000 x - x ^{2}} d x

对

5000 x - x^{2}

配方:

- (x - 2500)^{2} + 6250000

= \int \frac{1}{- ( x - 2500 ) ^{2} + 6250000} d x

使用换元积分法

= \int \frac{1}{- u ^{2} + 6250000} d u

使用换元积分法

= \int \frac{1}{2500 ( - v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{2500} \cdot \int \frac{1}{- v ^{2} + 1} d v

使用积分常见定则:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{2500} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

代回

= \frac{1}{2500} (\frac{ln \frac{x - 2500}{2500} + 1}{2} - \frac{ln \frac{x - 2500}{2500} - 1}{2})

化简

\frac{1}{2500} (\frac{ln \frac{x - 2500}{2500} + 1}{2} - \frac{ln \frac{x - 2500}{2500} - 1}{2}) : \frac{1}{5000} (ln \frac{x - 2500}{2500} + 1 - ln \frac{x - 2500}{2500} - 1)

= \frac{1}{5000} (ln \frac{x - 2500}{2500} + 1 - ln \frac{x - 2500}{2500} - 1)

解答补常数

= \frac{1}{5000} (ln \frac{x - 2500}{2500} + 1 - ln \frac{x - 2500}{2500} - 1) + C

作图

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