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受欢迎的微积分 >

integral of 1/(2+e^x)

解答

\int \frac{1}{2 + e ^{x}} d x

解答

\frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣ + C

求解步骤

\int \frac{1}{2 + e ^{x}} d x

使用换元积分法

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

将

\frac{1}{u ( 2 + u )}

用部份分式展开:

\frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

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

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

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

= \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣

u = e^{x}

代回

= \frac{1}{2} ln ∣ e^{x} ∣ - \frac{1}{2} ln ∣ e^{x} + 2 ∣

\frac{1}{2} ln ∣ e^{x} ∣ = \frac{1}{2} x

= \frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣

解答补常数

= \frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣ + C

作图

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