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

The integral of (x+1)e^{-2x} is -1/2 e^{-2x}x-3/4 e^{-2x}+C

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

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

- \frac{1}{2} e^{- 2 x} x - \frac{3}{4} e^{- 2 x} + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{1}{2} e^{- 2 x} d x = \frac{1}{4} e^{- 2 x}

= - \frac{1}{2} e^{- 2 x} (x + 1) - \frac{1}{4} e^{- 2 x}

Simplify

- \frac{1}{2} e^{- 2 x} (x + 1) - \frac{1}{4} e^{- 2 x} : - \frac{1}{2} e^{- 2 x} x - \frac{3}{4} e^{- 2 x}

= - \frac{1}{2} e^{- 2 x} x - \frac{3}{4} e^{- 2 x}

Add a constant to the solution

= - \frac{1}{2} e^{- 2 x} x - \frac{3}{4} e^{- 2 x} + C

f (x) = \frac{x ^{2} - 2 x - 3}{x}

\frac{\partial}{\partial x} (3 x^{2} - 12 y)

\frac{d}{d x} (\frac{18}{x ^{2} + 2})

\frac{4}{x - 4} + \frac{27}{2 x + 2}

x^{^{'}} - x = 0

What is the integral of (x+1)e^{-2x} ?
The integral of (x+1)e^{-2x} is -1/2 e^{-2x}x-3/4 e^{-2x}+C