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

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

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

integral

\frac{e ^{- x}}{2}

- \frac{1}{2} e^{- x} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - e^{u} d u

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

Take the constant out:

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

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

Use the common integral:

\int e^{u} d u = e^{u}

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

Substitute back

u = - x

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

Simplify

= - \frac{1}{2} e^{- x}

Add a constant to the solution

= - \frac{1}{2} e^{- x} + C

\frac{6}{( 1 + t ^{2} )}

a^{2} + x^{2}

ln (1 + x^{2})

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

\frac{1}{( x ^{4} - 1 )}

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