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受欢迎的微积分 >

integral of 2/(e^x+2)

解答

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

解答

x - ln ∣ e^{x} + 2 ∣ + C

求解步骤

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

提出常数:

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

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

使用换元积分法

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

将

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

用部份分式展开:

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

= 2 \cdot \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

= 2 (\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 ∣

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

u = e^{x}

代回

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

化简

2 (\frac{1}{2} ln ∣ e^{x} ∣ - \frac{1}{2} ln ∣ e^{x} + 2 ∣) : x - ln ∣ e^{x} + 2 ∣

= x - ln ∣ e^{x} + 2 ∣

解答补常数

= x - ln ∣ e^{x} + 2 ∣ + C

作图

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