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

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

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

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

- e^{- 2 x - 1} (x + 1) + C

Solution steps

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

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

= \frac{1}{e} \cdot \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}}

= \frac{1}{e} \cdot \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

= \frac{1}{e} (\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}

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

Simplify

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

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

Add a constant to the solution

= - e^{- 2 x - 1} (x + 1) + C

\int x (e^{2 x}) d x

\int (a x^{2} - b)^{5} x d x

arccot (x)

\frac{d}{d x} (2^{s i n (\frac{1}{x})})

x \to 0 lim (\frac{x}{sin ( x + π )})

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