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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add similar elements:

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

\frac{\partial}{\partial x} (e^{x} cos (y) + 2 x - 3 y^{2})

2 + t + \frac{1}{t}

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

\frac{d}{d x} (\frac{cosh ( x )}{sin ^{2} ( x )})

\frac{d}{d x} (5 cos (5 x))

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