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

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

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

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

- e^{- x} x + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= - e^{- x} x

Add a constant to the solution

= - e^{- x} x + C

\frac{d}{d x} (\frac{x + 2}{10 - 3 x})

\frac{d y}{d x}, a x^{3} - 3 b^{2} x y + cy (x)^{3} = 1

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

(\frac{x ^{3}}{x ^{8} + 3})^{2}

f (x) = ((2 x - 2) (3 x^{3} - x^{2} + 1))