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

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

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

Add a constant to the solution

= e^{- 2 x} x + C

\frac{d}{d x} (\frac{1}{2} x + \frac{1}{4} sin (2 x))

\int x^{2} e^{x^{3} - 1} d x

\frac{\partial}{\partial x} (x ln (x y - 9))

\frac{4 x ^{2} + 2 x + 8}{x}

x \to \frac{1}{2} lim (10 x^{2} - 9 x + 2)

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