What is the integral of (x-1)e^{-x-1} ?

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

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integral of (x-1)e^{-x-1}

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

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

Solution steps

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

Apply Integration By Parts

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

= - 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

n = 0 \sum \infty e

f (x) = x - \frac{1}{x}

\frac{d}{d x} (\frac{- 1}{( x - 2 ) ^{2}})

x \to - \infty lim (e^{x} + e^{- x})

\int 16 - 8 x^{2} + x^{4} d x

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