What is the integral of e^{tx}*1/θ e^{-x/θ} ?

The integral of e^{tx}*1/θ e^{-x/θ} is -(e^{-(x(-tθ+1))/θ})/(-tθ+1)+C

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integral of e^{tx}*1/θ e^{-x/θ}

\int e^{t x} \cdot \frac{1}{θ} e^{- \frac{x}{θ}} d x

- \frac{e ^{- \frac{x ( - tθ + 1 )}{θ}}}{- tθ + 1} + C

Solution steps

\int e^{t x} \frac{1}{θ} e^{- \frac{x}{θ}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{θ} \cdot \int e^{- \frac{x}{θ}} e^{t x} d x

Apply u-substitution

= \frac{1}{θ} \cdot \int - θ u^{- tθ} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{θ} (- θ \cdot \int u^{- tθ} d u)

Apply the Power Rule

= \frac{1}{θ} (- θ \frac{u ^{- tθ + 1}}{- tθ + 1})

Substitute back

u = e^{- \frac{x}{θ}}

= \frac{1}{θ} (- θ \frac{( e ^{- \frac{x}{θ}} ) ^{- tθ + 1}}{- tθ + 1})

Simplify

\frac{1}{θ} (- θ \frac{( e ^{- \frac{x}{θ}} ) ^{- tθ + 1}}{- tθ + 1}) : - \frac{e ^{- \frac{x ( - tθ + 1 )}{θ}}}{- tθ + 1}

= - \frac{e ^{- \frac{x ( - tθ + 1 )}{θ}}}{- tθ + 1}

Add a constant to the solution

= - \frac{e ^{- \frac{x ( - tθ + 1 )}{θ}}}{- tθ + 1} + C

What is the integral of e^{tx}*1/θ e^{-x/θ} ?
The integral of e^{tx}*1/θ e^{-x/θ} is -(e^{-(x(-tθ+1))/θ})/(-tθ+1)+C