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

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

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integral of e^{t-st}

\int e^{t - s t}

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = t - s t

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

Add a constant to the solution

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

g (x) = 10 x e^{3 - x^{2}}

7 t \cdot y^{^{'}} - 5 y = t

4 y^{^{''}} - y = 1, y (0) = 0, y^{^{'}} (0) = \frac{1}{2}

\frac{d y}{d x} = \frac{3 y - 2}{2 x + 1}

x \to 2 lim (\frac{e ^{3}}{x})

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