What is the integral of-e^{-st} ?

The integral of-e^{-st} is e^{-st} 1/s+C

Popular >

integral of-e^{-st}

\int - e^{- s t} d t

e^{- s t} \frac{1}{s} + C

Solution steps

\int - e^{- s t} d t

Take the constant out:

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

= - \int e^{- s t} d t

Apply u-substitution

: \int - \frac{e ^{u}}{s} d u

= - \int - \frac{e ^{u}}{s} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - s t

= - (- \frac{1}{s} e^{- s t})

Simplify

= e^{- s t} \frac{1}{s}

Add a constant to the solution

= e^{- s t} \frac{1}{s} + C

\int \frac{2}{e ^{t}} d t

\frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2} - 1})

\frac{d y}{d x} = \frac{2 x y}{( x ^{2} - y ^{2} )}

\frac{d y}{d x} - 5 y = - \frac{5}{2} x y^{3}

\int \frac{x}{x ^{2} + 6} d x