What is the integral of e^{-st}te^{4t} ?

The integral of e^{-st}te^{4t} is e^{-st+4t} 1/(-s+4)t-e^{-st+4t} 1/((-s+4)^2)+C

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integral of e^{-st}te^{4t}

\int e^{- s t} t e^{4 t} d t

e^{- s t + 4 t} \frac{1}{- s + 4} t - e^{- s t + 4 t} \frac{1}{( - s + 4 ) ^{2}} + C

Solution steps

\int e^{- s t} t e^{4 t} d t

Simplify

e^{- s t} e^{4 t} t : e^{- s t + 4 t} t

= \int e^{- s t + 4 t} t d t

Apply Integration By Parts

: e^{- s t + 4 t} \frac{1}{- s + 4} t - \int e^{- s t + 4 t} \frac{1}{- s + 4} d t

= e^{- s t + 4 t} \frac{1}{- s + 4} t - \int e^{- s t + 4 t} \frac{1}{- s + 4} d t

\int e^{- s t + 4 t} \frac{1}{- s + 4} d t = e^{- s t + 4 t} \frac{1}{( - s + 4 ) ^{2}}

= e^{- s t + 4 t} \frac{1}{- s + 4} t - e^{- s t + 4 t} \frac{1}{( - s + 4 ) ^{2}}

Add a constant to the solution

= e^{- s t + 4 t} \frac{1}{- s + 4} t - e^{- s t + 4 t} \frac{1}{( - s + 4 ) ^{2}} + C

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What is the integral of e^{-st}te^{4t} ?
The integral of e^{-st}te^{4t} is e^{-st+4t} 1/(-s+4)t-e^{-st+4t} 1/((-s+4)^2)+C