What is the integral of (e^{-x})/3 ?

The integral of (e^{-x})/3 is -1/3 e^{-x}+C

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integral of (e^{-x})/3

integral

\frac{e ^{- x}}{3}

- \frac{1}{3} e^{- x} + C

Solution steps

\int \frac{e ^{- x}}{3} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{- x} d x

Apply u-substitution

: \int - e^{u} d u

= \frac{1}{3} \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}{3} (- \int e^{u} d u)

Use the common integral:

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

= \frac{1}{3} (- e^{u})

Substitute back

u = - x

= \frac{1}{3} (- e^{- x})

Simplify

= - \frac{1}{3} e^{- x}

Add a constant to the solution

= - \frac{1}{3} e^{- x} + C

x (x^{2} + 4)^{3}

\frac{x ^{4}}{( 1 + x ^{2} )}

2 x^{- 5}

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

\frac{( z + 4 )}{( z ^{2} + 2 z + 5 )}

What is the integral of (e^{-x})/3 ?
The integral of (e^{-x})/3 is -1/3 e^{-x}+C