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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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sin^{2} (x^{2} - 1)

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