What is the integral of-1/5 e^{-5x} ?

The integral of-1/5 e^{-5x} is 1/25 e^{-5x}+C

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integral of-1/5 e^{-5x}

\int - \frac{1}{5} e^{- 5 x} d x

\frac{1}{25} e^{- 5 x} + C

Solution steps

\int - \frac{1}{5} e^{- 5 x} d x

Take the constant out:

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

= - \frac{1}{5} \cdot \int e^{- 5 x} d x

Apply u-substitution

: \int - \frac{1}{5} e^{u} d u

= - \frac{1}{5} \cdot \int - \frac{1}{5} e^{u} d u

Take the constant out:

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

= - \frac{1}{5} (- \frac{1}{5} \cdot \int e^{u} d u)

Use the common integral:

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

= - \frac{1}{5} (- \frac{1}{5} e^{u})

Substitute back

u = - 5 x

= - \frac{1}{5} (- \frac{1}{5} e^{- 5 x})

Simplify

- \frac{1}{5} (- \frac{1}{5} e^{- 5 x}) : \frac{1}{25} e^{- 5 x}

= \frac{1}{25} e^{- 5 x}

Add a constant to the solution

= \frac{1}{25} e^{- 5 x} + C

\int \frac{x}{2 x} d x

\frac{d}{d x} (x^{4} + 3 x^{3} - x^{2})

\frac{s + 1}{( s ^{2} - 4 s ) ( s + 5 )}

x \to 2 lim (ln (x^{2} - 4))

\frac{d}{d x} (\frac{e ^{x}}{x \cdot cos ( x )})

What is the integral of-1/5 e^{-5x} ?
The integral of-1/5 e^{-5x} is 1/25 e^{-5x}+C