What is the integral of (2^{2x}+2^{4x})/(2^{2x)+1} ?

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

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integral of (2^{2x}+2^{4x})/(2^{2x)+1}

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

\frac{1}{2 ln ( 2 )} ln ∣ 4^{x} + 1 ∣ + \frac{- ln ∣ 4 ^{x} + 1 ∣ + 4 ^{x} + 1}{2 ln ( 2 )} + C

Solution steps

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

Expand

\frac{2 ^{2 x} + 2 ^{4 x}}{2 ^{2 x} + 1} : \frac{2 ^{2 x}}{2 ^{2 x} + 1} + \frac{2 ^{4 x}}{2 ^{2 x} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 ^{2 x}}{2 ^{2 x} + 1} d x + \int \frac{2 ^{4 x}}{2 ^{2 x} + 1} d x

\int \frac{2 ^{2 x}}{2 ^{2 x} + 1} d x = \frac{1}{2 ln ( 2 )} ln ∣ 4^{x} + 1 ∣

\int \frac{2 ^{4 x}}{2 ^{2 x} + 1} d x = \frac{- ln ∣ 4 ^{x} + 1 ∣ + 4 ^{x} + 1}{2 ln ( 2 )}

= \frac{1}{2 ln ( 2 )} ln ∣ 4^{x} + 1 ∣ + \frac{- ln ∣ 4 ^{x} + 1 ∣ + 4 ^{x} + 1}{2 ln ( 2 )}

Add a constant to the solution

= \frac{1}{2 ln ( 2 )} ln ∣ 4^{x} + 1 ∣ + \frac{- ln ∣ 4 ^{x} + 1 ∣ + 4 ^{x} + 1}{2 ln ( 2 )} + C

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