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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2^{-x})/(ln(2))

\int \frac{2 ^{- x}}{ln ( 2 )} d x

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

Solution steps

\int \frac{2 ^{- x}}{ln ( 2 )} d x

Take the constant out:

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

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

Apply u-substitution

: \int - 2^{u} d u

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

Take the constant out:

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

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

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

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

Substitute back

u = - x

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{sin ( 3 ln ( x ) )}{x} d x

\int \frac{x}{x ^{2} - 7} d x

\int \frac{3 x ^{2}}{( 8 - 3 x ^{3} ) ^{5}} d x

\int - cos (x) x d x

\int \frac{8}{x ln ( 9 x )} d x

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