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

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

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

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

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

Solution steps

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

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

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

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

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

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

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

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

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

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

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

Add a constant to the solution

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

\int 1 - \frac{3}{14 - x} d x

\int \frac{3}{x + 3 + x} d x

\frac{d y}{d t} + y = 7 t

(7, - 2), (3, 5)

x \to 0 lim (\frac{7 x ^{2} - 6 x}{x})

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