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

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

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integral of e^{-t(s+1)}

\int e^{- t (s + 1)} d t

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

Solution steps

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

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

Add a constant to the solution

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

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

f (t) = 2^{t^{3}}

sin (x), (\frac{π}{4})

y^{^{'''}} - 6 y^{^{''}} = 4 - cos (x)

x \to 2 + lim (\frac{x}{x ^{2} - 4})

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