What is the integral from 0 to 1 of e^{-st}(2t+1) ?

The integral from 0 to 1 of e^{-st}(2t+1) is (-3e^{-s}+1)/s+(-2e^{-s}+2)/(s^2)

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integral from 0 to 1 of e^{-st}(2t+1)

\int_{0}^{1} e^{- s t} (2 t + 1) d t

\frac{- 3 e ^{- s} + 1}{s} + \frac{- 2 e ^{- s} + 2}{s ^{2}}

Solution steps

\int_{0}^{1} 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]_{0}^{1}

= [- e^{- s t} \frac{1}{s} (1 + 2 t) - \int - e^{- s t} \frac{2}{s} d t]_{0}^{1}

\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}}]_{0}^{1}

Compute the boundaries:

- e^{- s} \frac{3}{s} - e^{- s} \frac{2}{s ^{2}} + \frac{1}{s} + \frac{2}{s ^{2}}

= - e^{- s} \frac{3}{s} - e^{- s} \frac{2}{s ^{2}} + \frac{1}{s} + \frac{2}{s ^{2}}

Simplify

= \frac{- 3 e ^{- s} + 1}{s} + \frac{- 2 e ^{- s} + 2}{s ^{2}}

\int_{0}^{\frac{π}{2}} sin (x^{2}) d x

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\int_{0}^{\frac{π}{2}} x \cdot cos (x) d x

\int_{1}^{2} \frac{x + 1}{x ^{2} + 2 x} d x

\int_{- 2}^{- 1} 2 x^{- 4} d x

What is the integral from 0 to 1 of e^{-st}(2t+1) ?
The integral from 0 to 1 of e^{-st}(2t+1) is (-3e^{-s}+1)/s+(-2e^{-s}+2)/(s^2)