What is the integral of (2t+1)e^{-st} ?

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

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integral of (2t+1)e^{-st}

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Add a constant to the solution

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

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