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受欢迎的微积分 >

integral of (e^x)/((e^x-6)(e^x+1))

解答

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

解答

- \frac{1}{7} ln ∣ e^{x} + 1 ∣ + \frac{1}{7} ln ∣ e^{x} - 6 ∣ + C

求解步骤

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

乘开

(e^{x} - 6) (e^{x} + 1) : e^{2 x} - 5 e^{x} - 6

= \int \frac{e ^{x}}{e ^{2 x} - 5 e ^{x} - 6} d x

使用换元积分法

= \int \frac{1}{u ^{2} - 5 u - 6} d u

将

\frac{1}{u ^{2} - 5 u - 6}

用部份分式展开:

- \frac{1}{7 ( u + 1 )} + \frac{1}{7 ( u - 6 )}

= \int - \frac{1}{7 ( u + 1 )} + \frac{1}{7 ( u - 6 )} 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 \frac{1}{7 ( u + 1 )} d u + \int \frac{1}{7 ( u - 6 )} d u

\int \frac{1}{7 ( u + 1 )} d u = \frac{1}{7} ln ∣ u + 1 ∣

\int \frac{1}{7 ( u - 6 )} d u = \frac{1}{7} ln ∣ u - 6 ∣

= - \frac{1}{7} ln ∣ u + 1 ∣ + \frac{1}{7} ln ∣ u - 6 ∣

u = e^{x}

代回

= - \frac{1}{7} ln ∣ e^{x} + 1 ∣ + \frac{1}{7} ln ∣ e^{x} - 6 ∣

解答补常数

= - \frac{1}{7} ln ∣ e^{x} + 1 ∣ + \frac{1}{7} ln ∣ e^{x} - 6 ∣ + C

作图

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