What is the integral of (e^{x-1+ln(x-1)})/(e^{2x)} ?

The integral of (e^{x-1+ln(x-1)})/(e^{2x)} is -e^{-x-1}x+C

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integral of (e^{x-1+ln(x-1)})/(e^{2x)}

\int \frac{e ^{x - 1 + l n (x - 1)}}{e ^{2 x}} d x

- e^{- x - 1} x + C

Solution steps

\int \frac{e ^{x - 1 + l n (x - 1)}}{e ^{2 x}} d x

Simplify

\frac{e ^{x - 1 + l n (x - 1)}}{e ^{2 x}} : e^{- x - 1} (x - 1)

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

Apply Integration By Parts

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

\int - e^{- x - 1} d x = e^{- x - 1}

= - e^{- x - 1} (x - 1) - e^{- x - 1}

Simplify

- e^{- x - 1} (x - 1) - e^{- x - 1} : - e^{- x - 1} x

= - e^{- x - 1} x

Add a constant to the solution

= - e^{- x - 1} x + C

What is the integral of (e^{x-1+ln(x-1)})/(e^{2x)} ?
The integral of (e^{x-1+ln(x-1)})/(e^{2x)} is -e^{-x-1}x+C