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受欢迎的微积分 >

integral of (e^{2x}+4)^{-1}

解答

\int (e^{2 x} + 4)^{- 1} d x

解答

\frac{1}{8} (- ln e^{2 x} + 4 + 2 x) + C

求解步骤

\int (e^{2 x} + 4)^{- 1} d x

使用指数法则:

a^{- 1} = \frac{1}{a}

= \int \frac{1}{e ^{2 x} + 4} d x

使用换元积分法

= \int \frac{1}{2 u ( u - 4 )} 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{1}{u ( u - 4 )} d u

将

\frac{1}{u ( u - 4 )}

用部份分式展开:

- \frac{1}{4 u} + \frac{1}{4 ( u - 4 )}

= \frac{1}{2} \cdot \int - \frac{1}{4 u} + \frac{1}{4 ( u - 4 )} 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 \frac{1}{4 u} d u + \int \frac{1}{4 ( u - 4 )} d u)

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u - 4 )} d u = \frac{1}{4} ln ∣ u - 4 ∣

= \frac{1}{2} (- \frac{1}{4} ln ∣ u ∣ + \frac{1}{4} ln ∣ u - 4 ∣)

u = e^{2 x} + 4

代回

= \frac{1}{2} (- \frac{1}{4} ln e^{2 x} + 4 + \frac{1}{4} ln e^{2 x} + 4 - 4)

化简

\frac{1}{2} (- \frac{1}{4} ln e^{2 x} + 4 + \frac{1}{4} ln e^{2 x} + 4 - 4) : \frac{1}{8} (- ln e^{2 x} + 4 + 2 x)

= \frac{1}{8} (- ln e^{2 x} + 4 + 2 x)

解答补常数

= \frac{1}{8} (- ln e^{2 x} + 4 + 2 x) + C

作图

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