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

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

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integral of 2^{1-x}

\int 2^{1 - x} d x

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

Solution steps

\int 2^{1 - x} d x

Apply u-substitution

: \int - 2^{u} d u

= \int - 2^{u} d u

Take the constant out:

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

= - \int 2^{u} d u

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

= - \frac{2 ^{u}}{ln ( 2 )}

Substitute back

u = 1 - x

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{n}{4 ^{n}}

\frac{d}{d x} (6 x - 4)

y^{^{''}} + y^{^{'}} + y = x e^{x} + e^{- x} (1 + 2 x)

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

f (x) = 8 sin^{4} (x)

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