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

The integral of e^{-x}-xe^{-x} is e^{-x}x+C

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

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

e^{- x} x + C

Solution steps

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

Apply u-substitution

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int e^{u} d u = e^{u}

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

= - e^{u} - (e^{u} u - e^{u})

Substitute back

u = - x

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

Simplify

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

= e^{- x} x

Add a constant to the solution

= e^{- x} x + C

What is the integral of e^{-x}-xe^{-x} ?
The integral of e^{-x}-xe^{-x} is e^{-x}x+C