What is the integral of 4^{1-x} ?

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

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integral of 4^{1-x}

\int 4^{1 - x} d x

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

Solution steps

\int 4^{1 - x} d x

Apply u-substitution

: \int - 4^{u} d u

= \int - 4^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int 4^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - \frac{4 ^{u}}{ln ( 4 )}

Substitute back

u = 1 - x

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

Simplify

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

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

Add a constant to the solution

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

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