What is the integral of ((4^{1+x}-8^{1+x})/(2^{2x)}) ?

The integral of ((4^{1+x}-8^{1+x})/(2^{2x)}) is 4x-(2^{x+3})/(ln(2))+C

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integral of ((4^{1+x}-8^{1+x})/(2^{2x)})

\int (\frac{4 ^{1 + x} - 8 ^{1 + x}}{2 ^{2 x}}) d x

4 x - \frac{2 ^{x + 3}}{ln ( 2 )} + C

Solution steps

\int \frac{4 ^{1 + x} - 8 ^{1 + x}}{2 ^{2 x}} d x

Expand

\frac{4 ^{1 + x} - 8 ^{1 + x}}{2 ^{2 x}} : 4 - \frac{8 ^{1 + x}}{4 ^{x}}

= \int 4 - \frac{8 ^{1 + x}}{4 ^{x}} d 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 4 d x - \int \frac{8 ^{1 + x}}{4 ^{x}} d x

\int 4 d x = 4 x

\int \frac{8 ^{1 + x}}{4 ^{x}} d x = \frac{2 ^{x + 3}}{ln ( 2 )}

= 4 x - \frac{2 ^{x + 3}}{ln ( 2 )}

Add a constant to the solution

= 4 x - \frac{2 ^{x + 3}}{ln ( 2 )} + C

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What is the integral of ((4^{1+x}-8^{1+x})/(2^{2x)}) ?
The integral of ((4^{1+x}-8^{1+x})/(2^{2x)}) is 4x-(2^{x+3})/(ln(2))+C