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受欢迎的微积分 >

integral of 2^{-x}tanh(2^{1-x})

解答

\int 2^{- x} tanh (2^{1 - x}) d x

解答

- \frac{1}{2 ln ( 2 )} ln cosh (2^{1 - x}) + C

求解步骤

\int 2^{- x} tanh (2^{1 - x}) d x

使用换元积分法

= \int - \frac{2 ^{\frac{l n ( u ) - l n ( 2 )}{l n ( 2 )}} tanh ( u )}{ln ( 2 ) u} d u

提出常数:

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

= - \frac{1}{ln ( 2 )} \cdot \int \frac{2 ^{\frac{l n ( u ) - l n ( 2 )}{l n ( 2 )}} tanh ( u )}{u} d u

使用换元积分法

= - \frac{1}{ln ( 2 )} \cdot \int tanh (2 v) d v

使用换元积分法

= - \frac{1}{ln ( 2 )} \cdot \int tanh (w) \frac{1}{2} d w

提出常数:

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

= - \frac{1}{ln ( 2 )} \cdot \frac{1}{2} \cdot \int tanh (w) d w

使用三角恒等式改写

= - \frac{1}{ln ( 2 )} \cdot \frac{1}{2} \cdot \int \frac{sinh ( w )}{cosh ( w )} d w

使用换元积分法

= - \frac{1}{ln ( 2 )} \cdot \frac{1}{2} \cdot \int \frac{1}{t} d t

使用积分常见定则:

\int \frac{1}{t} d t = ln (∣ t ∣)

= - \frac{1}{ln ( 2 )} \cdot \frac{1}{2} ln ∣ t ∣

代回

= - \frac{1}{ln ( 2 )} \cdot \frac{1}{2} ln cosh (2 \cdot 2^{\frac{l n ( 2 ^{1 - x} ) - l n ( 2 )}{l n ( 2 )}})

化简

- \frac{1}{ln ( 2 )} \cdot \frac{1}{2} ln cosh (2 \cdot 2^{\frac{l n ( 2 ^{1 - x} ) - l n ( 2 )}{l n ( 2 )}}) : - \frac{1}{2 ln ( 2 )} ln cosh (2^{1 - x})

= - \frac{1}{2 ln ( 2 )} ln cosh (2^{1 - x})

解答补常数

= - \frac{1}{2 ln ( 2 )} ln cosh (2^{1 - x}) + C

作图

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