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

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

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

\int \frac{e ^{u}}{4 + e ^{2 u}} d u

\frac{1}{2} arctan (\frac{e ^{u}}{2}) + C

Solution steps

\int \frac{e ^{u}}{4 + e ^{2 u}} d u

Apply u-substitution

: \int \frac{1}{4 + v ^{2}} d v

= \int \frac{1}{4 + v ^{2}} d v

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{2} arctan (w)

Substitute back

= \frac{1}{2} arctan (\frac{e ^{u}}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{e ^{u}}{2}) + C

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