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

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

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

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

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

Solution steps

\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}})

= \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

= \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}

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

Add a constant to the solution

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

y^{^{'}} = - 2 x y

\frac{d y}{d x} = cot (1 - 2 x^{2})

\int \frac{1}{z ^{3} z ^{2} - 64} d z

\int \frac{9}{x + 2} d x

x \to 4 - lim (\frac{3}{x - 4})

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