What is the integral of ((1+e^{2x})/(4e^x)) ?

The integral of ((1+e^{2x})/(4e^x)) is 1/4 (-1/(e^x)+e^x)+C

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integral of ((1+e^{2x})/(4e^x))

\int (\frac{1 + e ^{2 x}}{4 e ^{x}}) d x

\frac{1}{4} (- \frac{1}{e ^{x}} + e^{x}) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{4} \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}})

= \frac{1}{4} \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

= \frac{1}{4} (\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}{4} (- \frac{1}{e ^{x}} + e^{x})

Add a constant to the solution

= \frac{1}{4} (- \frac{1}{e ^{x}} + e^{x}) + C

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What is the integral of ((1+e^{2x})/(4e^x)) ?
The integral of ((1+e^{2x})/(4e^x)) is 1/4 (-1/(e^x)+e^x)+C