What is the integral of (2^x)/(1-4^x) ?

The integral of (2^x)/(1-4^x) is (ln|2^{-x}+1|-ln|2^{-x}-1|)/(2ln(2))+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\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

Apply u-substitution

: \int - \frac{2}{1 - 4 v ^{2}} d v

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

Take the constant out:

\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)

Apply Integral Substitution

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

Take the constant out:

\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)

Use the common integral:

\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}))

Substitute back

= - \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}

Simplify

- \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 )}

Add a constant to the solution

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

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What is the integral of (2^x)/(1-4^x) ?
The integral of (2^x)/(1-4^x) is (ln|2^{-x}+1|-ln|2^{-x}-1|)/(2ln(2))+C