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

The integral of (2x+1)e^{-x} is -2e^{-x}x-3e^{-x}+C

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int - 2 e^{- x} d x = 2 e^{- x}

= - e^{- x} (2 x + 1) - 2 e^{- x}

Simplify

- e^{- x} (2 x + 1) - 2 e^{- x} : - 2 e^{- x} x - 3 e^{- x}

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

Add a constant to the solution

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

x \to 3 lim (x (x - 3))

y^{^{'''}} - 2 y^{^{'}} - 4 y = 0

t + 1^{2}

\int_{0}^{11} 2 π y (\frac{y ^{2}}{2}) d y

\frac{s}{( s + 4 )}

What is the integral of (2x+1)e^{-x} ?
The integral of (2x+1)e^{-x} is -2e^{-x}x-3e^{-x}+C