What is the integral of e^{(a-s)t} ?

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

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integral of e^{(a-s)t}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = (a - s) t

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

Add a constant to the solution

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

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\int 3 sin (2 x) d x

16 \frac{d ^{4} y}{d x ^{4}} + 48 \frac{d ^{2} y}{d x ^{2}} + 36 y = 0

x, x^{2} - 4 x + 3

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

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