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

The integral of e^{-st}e^{at} is 1/(-s+a)e^{-st+at}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{-st}e^{at}

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

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

Solution steps

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

Simplify

e^{- s t} e^{a t} : e^{- s t + a t}

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{- s + a} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{- s + a} e^{u}

Substitute back

u = - s t + a t

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

Add a constant to the solution

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

\int 2 x^{4} + 5 x + 1 d x

\frac{y ^{^{'}}}{y} = 5

n = 1 \sum \infty \frac{1}{5 ^{n + 2}}

(\frac{1}{2})^{9}

\int \frac{3 x + 1}{x ( x + 1 )} d x

What is the integral of e^{-st}e^{at} ?
The integral of e^{-st}e^{at} is 1/(-s+a)e^{-st+at}+C