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

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

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integral of 2^{2-2x}

\int 2^{2 - 2 x} d x

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

Solution steps

\int 2^{2 - 2 x} d x

2^{2 - 2 x} = 2^{2} \cdot 2^{- 2 x}

= \int 2^{2} \cdot 2^{- 2 x} d x

Take the constant out:

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

= 2^{2} \cdot \int 2^{- 2 x} d x

Simplify

= 4 \cdot \int 2^{- 2 x} d x

Apply u-substitution

: \int - 2^{u - 1} d u

= 4 \cdot \int - 2^{u - 1} d u

Take the constant out:

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

= 4 (- \int 2^{u - 1} d u)

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= 4 (- \int 2^{u} \cdot 2^{- 1} d u)

Take the constant out:

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

= 4 (- 2^{- 1} \cdot \int 2^{u} d u)

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= 4 (- \frac{1}{2} \cdot \int 2^{u} d u)

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

= 4 (- \frac{1}{2} \cdot \frac{2 ^{u}}{ln ( 2 )})

Substitute back

u = - 2 x

= 4 (- \frac{1}{2} \cdot \frac{2 ^{- 2 x}}{ln ( 2 )})

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{1} (e^{x} - x) d x

\frac{1}{x - 1}, 24

x = y (2 - y), x = - y

x \to 343 lim (x^{\frac{2}{3}})

\int_{- 2}^{3} \frac{41}{x ^{4}} d x

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