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受欢迎的微积分 >

integral of 2/(e^{2x)+2e^x+15}

解答

\int \frac{2}{e ^{2 x} + 2 e ^{x} + 15} d x

解答

2 (\frac{x}{15} + \frac{1}{15} (- \frac{1}{2} ln 2 e^{x} + e^{2 x} + 15 - \frac{1}{14} arctan (\frac{1}{14} (e^{x} + 1)))) + C

求解步骤

\int \frac{2}{e ^{2 x} + 2 e ^{x} + 15} 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 ^{2 x} + 2 e ^{x} + 15} d x

使用换元积分法

= 2 \cdot \int \frac{1}{u ( u ^{2} + 2 u + 15 )} d u

对

u^{2} + 2 u + 15

配方:

(u + 1)^{2} + 14

= 2 \cdot \int \frac{1}{u ( ( u + 1 ) ^{2} + 14 )} d u

使用换元积分法

= 2 \cdot \int \frac{1}{( v - 1 ) ( v ^{2} + 14 )} d v

将

\frac{1}{( v - 1 ) ( v ^{2} + 14 )}

用部份分式展开:

\frac{1}{15 ( v - 1 )} + \frac{- v - 1}{15 ( v ^{2} + 14 )}

= 2 \cdot \int \frac{1}{15 ( v - 1 )} + \frac{- v - 1}{15 ( v ^{2} + 14 )} d v

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}{15 ( v - 1 )} d v + \int \frac{- v - 1}{15 ( v ^{2} + 14 )} d v)

\int \frac{1}{15 ( v - 1 )} d v = \frac{1}{15} ln ∣ v - 1 ∣

\int \frac{- v - 1}{15 ( v ^{2} + 14 )} d v = \frac{1}{15} (- \frac{1}{2} ln v^{2} + 14 - \frac{1}{14} arctan (\frac{v}{14}))

= 2 (\frac{1}{15} ln ∣ v - 1 ∣ + \frac{1}{15} (- \frac{1}{2} ln v^{2} + 14 - \frac{1}{14} arctan (\frac{v}{14})))

代回

= 2 (\frac{1}{15} ln ∣ e^{x} + 1 - 1 ∣ + \frac{1}{15} (- \frac{1}{2} ln (e^{x} + 1)^{2} + 14 - \frac{1}{14} arctan (\frac{e ^{x} + 1}{14})))

化简

2 (\frac{1}{15} ln ∣ e^{x} + 1 - 1 ∣ + \frac{1}{15} (- \frac{1}{2} ln (e^{x} + 1)^{2} + 14 - \frac{1}{14} arctan (\frac{e ^{x} + 1}{14}))) : 2 (\frac{x}{15} + \frac{1}{15} (- \frac{1}{2} ln 2 e^{x} + e^{2 x} + 15 - \frac{1}{14} arctan (\frac{1}{14} (e^{x} + 1))))

= 2 (\frac{x}{15} + \frac{1}{15} (- \frac{1}{2} ln 2 e^{x} + e^{2 x} + 15 - \frac{1}{14} arctan (\frac{1}{14} (e^{x} + 1))))

解答补常数

= 2 (\frac{x}{15} + \frac{1}{15} (- \frac{1}{2} ln 2 e^{x} + e^{2 x} + 15 - \frac{1}{14} arctan (\frac{1}{14} (e^{x} + 1)))) + C

作图

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