What is the integral of (4^x+5*16^x)/(1+16^x) ?

The integral of (4^x+5*16^x)/(1+16^x) is -(arctan(4^{-x}))/(2ln(2))+5/(4ln(2))ln|1+16^x|+C

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integral of (4^x+5*16^x)/(1+16^x)

\int \frac{4 ^{x} + 5 \cdot 1 6 ^{x}}{1 + 1 6 ^{x}} d x

- \frac{arctan ( 4 ^{- x} )}{2 ln ( 2 )} + \frac{5}{4 ln ( 2 )} ln ∣ 1 + 1 6^{x} ∣ + C

Solution steps

\int \frac{4 ^{x} + 5 \cdot 1 6 ^{x}}{1 + 1 6 ^{x}} d x

Expand

\frac{4 ^{x} + 5 \cdot 1 6 ^{x}}{1 + 1 6 ^{x}} : \frac{4 ^{x}}{1 + 1 6 ^{x}} + \frac{5 \cdot 1 6 ^{x}}{1 + 1 6 ^{x}}

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{4 ^{x}}{1 + 1 6 ^{x}} d x + \int \frac{5 \cdot 1 6 ^{x}}{1 + 1 6 ^{x}} d x

\int \frac{4 ^{x}}{1 + 1 6 ^{x}} d x = - \frac{arctan ( 4 ^{- x} )}{2 ln ( 2 )}

\int \frac{5 \cdot 1 6 ^{x}}{1 + 1 6 ^{x}} d x = \frac{5}{4 ln ( 2 )} ln ∣ 1 + 1 6^{x} ∣

= - \frac{arctan ( 4 ^{- x} )}{2 ln ( 2 )} + \frac{5}{4 ln ( 2 )} ln ∣ 1 + 1 6^{x} ∣

Add a constant to the solution

= - \frac{arctan ( 4 ^{- x} )}{2 ln ( 2 )} + \frac{5}{4 ln ( 2 )} ln ∣ 1 + 1 6^{x} ∣ + C

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What is the integral of (4^x+5*16^x)/(1+16^x) ?
The integral of (4^x+5*16^x)/(1+16^x) is -(arctan(4^{-x}))/(2ln(2))+5/(4ln(2))ln|1+16^x|+C