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

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

Solution steps

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

Expand

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

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

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

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

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

Add similar elements:

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

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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