What is the integral from 0 to 1 of e^{-st}(te^t) ?

The integral from 0 to 1 of e^{-st}(te^t) is (-se^{-s+1}+1)/((-s+1)^2)

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integral from 0 to 1 of e^{-st}(te^t)

\int_{0}^{1} e^{- s t} (t e^{t}) d t

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

Solution steps

\int_{0}^{1} e^{- s t} t e^{t} d t

Simplify

e^{- s t} e^{t} : e^{- s t + t}

= \int_{0}^{1} e^{- s t + t} t d t

Factor

- s t + t : t (- s + 1)

= \int_{0}^{1} e^{t (- s + 1)} t d t

Apply u-substitution

: \int_{0}^{- s + 1} \frac{e ^{u} u}{( - s + 1 ) ^{2}} d u

= \int_{0}^{- s + 1} \frac{e ^{u} u}{( - s + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{( - s + 1 ) ^{2}} \cdot \int_{0}^{- s + 1} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- s + 1}

= \frac{1}{( - s + 1 ) ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- s + 1}

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

= \frac{1}{( - s + 1 ) ^{2}} [e^{u} u - e^{u}]_{0}^{- s + 1}

Simplify

\frac{1}{( - s + 1 ) ^{2}} [e^{u} u - e^{u}]_{0}^{- s + 1} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s + 1}}{( - s + 1 ) ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- s + 1}}{( - s + 1 ) ^{2}}

Compute the boundaries:

- s e^{- s + 1} + 1

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

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What is the integral from 0 to 1 of e^{-st}(te^t) ?
The integral from 0 to 1 of e^{-st}(te^t) is (-se^{-s+1}+1)/((-s+1)^2)