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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

= - 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

y = \frac{1}{x ^{2}}, y = x, y = \frac{1}{8} x

\frac{d}{d x} (50 e^{- 0.5 x} - (\frac{5}{x}))

\frac{\partial}{\partial x} (x + e^{4 y})

\int \frac{( 3 x + 4 )}{( x ^{2} + 4 ) ( 3 - x )} d x

- (x)^{2} 1

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