What is the integral of e^{-5r} ?

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

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integral of e^{-5r}

\int e^{- 5 r} d r

- \frac{1}{5} e^{- 5 r} + C

Solution steps

\int e^{- 5 r} d r

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = - 5 r

= - \frac{1}{5} e^{- 5 r}

Add a constant to the solution

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

y = e^{2 x - x^{2}}

\frac{d y}{d x} = \frac{8 x}{7 y}

x \to - \infty lim (\frac{0}{x})

\int ln (\frac{1 + x}{1 - x}) d x

\frac{\partial}{\partial x} (\frac{x + y ^{2}}{z})