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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

- y = x y^{^{'}}

x y^{^{'}} + (x + 1) y = 6

(\frac{d y}{d x})^{2} + 2 e^{x} (\frac{d y}{d x}) + e^{2 x} = 0

\frac{\partial}{\partial x} (2 x y e^{x y^{2}})

\int \frac{1}{- 1} d x

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