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

The integral of e^{-t(s-1)} is e^{t(1-s)}t-(1-s) 1/((1-s)^2)(e^{t(1-s)}t(1-s)-e^{t(1-s)})+C

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

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

e^{t (1 - s)} t - (1 - s) \frac{1}{( 1 - s ) ^{2}} (e^{t (1 - s)} t (1 - s) - e^{t (1 - s)}) + C

Solution steps

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

Factor

- 1 + s : - (1 - s)

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

Simplify

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

Apply Integration By Parts

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

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

\int e^{t (1 - s)} (1 - s) t d t = (1 - s) \frac{1}{( 1 - s ) ^{2}} (e^{t (1 - s)} t (1 - s) - e^{t (1 - s)})

= e^{t (1 - s)} t - (1 - s) \frac{1}{( 1 - s ) ^{2}} (e^{t (1 - s)} t (1 - s) - e^{t (1 - s)})

Add a constant to the solution

= e^{t (1 - s)} t - (1 - s) \frac{1}{( 1 - s ) ^{2}} (e^{t (1 - s)} t (1 - s) - e^{t (1 - s)}) + C

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What is the integral of e^{-t(s-1)} ?
The integral of e^{-t(s-1)} is e^{t(1-s)}t-(1-s) 1/((1-s)^2)(e^{t(1-s)}t(1-s)-e^{t(1-s)})+C