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受欢迎的微积分 >

integral of (2^x)/(1-4^x)

解答

\int \frac{2 ^{x}}{1 - 4 ^{x}} d x

解答

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

求解步骤

\int \frac{2 ^{x}}{1 - 4 ^{x}} d x

使用换元积分法

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

使用换元积分法

= - \frac{1}{ln ( 2 )} \cdot \int - \frac{2}{1 - 4 v ^{2}} d v

提出常数:

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

= - \frac{1}{ln ( 2 )} (- 2 \cdot \int \frac{1}{1 - 4 v ^{2}} d v)

使用换元积分法

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

使用积分常见定则:

\int \frac{1}{1 - w ^{2}} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

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

代回

= - \frac{1}{ln ( 2 )} - 2 \cdot \frac{1}{2} \frac{ln 2 \cdot 2 ^{\frac{- l n ( - ( 1 - 4 ^{x} ) + 1 ) - 2 l n ( 2 )}{2 l n ( 2 )}} + 1}{2} - \frac{ln 2 \cdot 2 ^{\frac{- l n ( - ( 1 - 4 ^{x} ) + 1 ) - 2 l n ( 2 )}{2 l n ( 2 )}} - 1}{2}

化简

- \frac{1}{ln ( 2 )} - 2 \cdot \frac{1}{2} \frac{ln 2 \cdot 2 ^{\frac{- l n ( - ( 1 - 4 ^{x} ) + 1 ) - 2 l n ( 2 )}{2 l n ( 2 )}} + 1}{2} - \frac{ln 2 \cdot 2 ^{\frac{- l n ( - ( 1 - 4 ^{x} ) + 1 ) - 2 l n ( 2 )}{2 l n ( 2 )}} - 1}{2} : \frac{ln ∣ 2 ^{- x} + 1 ∣ - ln ∣ 2 ^{- x} - 1 ∣}{2 ln ( 2 )}

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

解答补常数

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

作图

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