What is the integral of e^{-st}(t-10) ?

The integral of e^{-st}(t-10) is -e^{-st} 1/s (-10+t)-e^{-st} 1/(s^2)+C

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integral of e^{-st}(t-10)

\int e^{- s t} (t - 10) d t

- e^{- s t} \frac{1}{s} (- 10 + t) - e^{- s t} \frac{1}{s ^{2}} + C

Solution steps

\int e^{- s t} (t - 10) d t

Apply Integration By Parts

: - e^{- s t} \frac{1}{s} (- 10 + t) - \int - e^{- s t} \frac{1}{s} d t

= - e^{- s t} \frac{1}{s} (- 10 + 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} (- 10 + t) - e^{- s t} \frac{1}{s ^{2}}

Add a constant to the solution

= - e^{- s t} \frac{1}{s} (- 10 + t) - e^{- s t} \frac{1}{s ^{2}} + C

sin (4 t - 5)

\frac{\partial}{\partial x} (x y (1 - x^{2} - y^{2}))

\frac{d}{d x} (\frac{2}{x - 4})

\int \frac{x}{4 x + 1} d x

(6 x y - 2 y^{2} + 1) d x + (3 x^{2} - 4 x y) d y = 0

What is the integral of e^{-st}(t-10) ?
The integral of e^{-st}(t-10) is -e^{-st} 1/s (-10+t)-e^{-st} 1/(s^2)+C