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

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

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

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

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

Solution steps

\int (x + 1) e^{- x} 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 - 2 e^{- x}

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

Add a constant to the solution

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

\int x e^{0.2 x} d x

f (x) = x^{3} + 3 x

x^{\frac{5}{7}}

f (x) = \frac{x + 9}{x - 9}

t^{2} y^{^{''}} - 5 t y^{^{'}} + 9 y = 0

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