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受欢迎的微积分 >

integral of 1/(36e^{-5x)+e^{5x}}

解答

\int \frac{1}{36 e ^{- 5 x} + e ^{5 x}} d x

解答

- \frac{1}{30} arctan (6 e^{- 5 x}) + C

求解步骤

\int \frac{1}{36 e ^{- 5 x} + e ^{5 x}} d x

使用换元积分法

= \int - \frac{1}{5 ( 36 u ^{2} + 1 )} d u

提出常数:

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

= - \frac{1}{5} \cdot \int \frac{1}{36 u ^{2} + 1} d u

使用换元积分法

= - \frac{1}{5} \cdot \int \frac{1}{6 ( v ^{2} + 1 )} d v

提出常数:

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

= - \frac{1}{5} \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

使用积分常见定则:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= - \frac{1}{5} \cdot \frac{1}{6} arctan (v)

代回

= - \frac{1}{5} \cdot \frac{1}{6} arctan (6 e^{- 5 x})

化简

- \frac{1}{5} \cdot \frac{1}{6} arctan (6 e^{- 5 x}) : - \frac{1}{30} arctan (6 e^{- 5 x})

= - \frac{1}{30} arctan (6 e^{- 5 x})

解答补常数

= - \frac{1}{30} arctan (6 e^{- 5 x}) + C

作图

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