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

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

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integral of 2^{-x}

integral

2^{- x}

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

Solution steps

\int 2^{- 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 = - x

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

Add a constant to the solution

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

sin (y^{2})

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

e^{- 3} x

\frac{1}{( x - 2 )}

\frac{1}{x ^{2} + a ^{2}}