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

The integral of (x^2+x)e^{-x} is -e^{-x}x^2-3e^{-x}x-3e^{-x}+C

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

\int (x^{2} + x) e^{- x} d x

- e^{- x} x^{2} - 3 e^{- x} x - 3 e^{- x} + C

Solution steps

\int (x^{2} + x) e^{- x} d x

Apply u-substitution

: \int - e^{u} (u^{2} - u) d u

= \int - e^{u} (u^{2} - u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int e^{u} (u^{2} - u) d u

Expand

e^{u} (u^{2} - u) : e^{u} u^{2} - e^{u} u

= - \int e^{u} u^{2} - 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} u^{2} d u - \int e^{u} u d u)

\int e^{u} u^{2} d u = u^{2} e^{u} - 2 (e^{u} u - e^{u})

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

= - (u^{2} e^{u} - 2 (e^{u} u - e^{u}) - (e^{u} u - e^{u}))

Substitute back

u = - x

= - ((- x)^{2} e^{- x} - 2 (e^{- x} (- x) - e^{- x}) - (e^{- x} (- x) - e^{- x}))

Simplify

- ((- x)^{2} e^{- x} - 2 (e^{- x} (- x) - e^{- x}) - (e^{- x} (- x) - e^{- x})) : - e^{- x} x^{2} - 3 e^{- x} x - 3 e^{- x}

= - e^{- x} x^{2} - 3 e^{- x} x - 3 e^{- x}

Add a constant to the solution

= - e^{- x} x^{2} - 3 e^{- x} x - 3 e^{- x} + C

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What is the integral of (x^2+x)e^{-x} ?
The integral of (x^2+x)e^{-x} is -e^{-x}x^2-3e^{-x}x-3e^{-x}+C