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

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

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

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

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

Solution steps

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

Expand

\frac{1 - 2 x}{e ^{- 2 x}} : \frac{1}{e ^{- 2 x}} - \frac{2 x}{e ^{- 2 x}}

Apply the Sum Rule:

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int (\frac{1}{2} + \frac{1}{x}) d x

\frac{d y}{d x} = \frac{( x - 4 )}{y + 3}

(1, - 7), (2, - 1)

\int e^{\frac{x}{3}} d x

ln (x + 1 + x^{2} + 2 x), 1

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