What is the integral of 16*2^{-x}-1 ?

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

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integral of 16*2^{-x}-1

\int 16 \cdot 2^{- x} - 1 d x

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

Solution steps

\int 16 \cdot 2^{- x} - 1 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 16 \cdot 2^{- x} d x - \int 1 d x

\int 16 \cdot 2^{- x} d x = - \frac{2 ^{- x + 4}}{ln ( 2 )}

\int 1 d x = x

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

Add a constant to the solution

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

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