What is the integral of e^{tx}xe^{-x} ?

The integral of e^{tx}xe^{-x} is -e^{-x+tx} 1/(1-t)x+e^{x(t-1)} 1/((1-t)(t-1))+C

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integral of e^{tx}xe^{-x}

\int e^{t x} x e^{- x} d x

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

Solution steps

\int e^{t x} x e^{- x} d x

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

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