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

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

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integral of 2^{-2x}

\int 2^{- 2 x} d x

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

Solution steps

\int 2^{- 2 x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Simplify

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

= - \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

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

Apply exponent rule:

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

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

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

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

Substitute back

u = - 2 x

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

Simplify

- \frac{1}{2} \cdot \frac{2 ^{- 2 x}}{ln ( 2 )} : - \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 2^{-2x} ?
The integral of 2^{-2x} is -(2^{-2x-1})/(ln(2))+C