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

The integral of (2e^{2x})/(2+e^{2x)} is ln|2+e^{2x}|+C

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

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

ln 2 + e^{2 x} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= 2 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 + e^{2 x}

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

Simplify

2 \cdot \frac{1}{2} ln 2 + e^{2 x} : ln 2 + e^{2 x}

= ln 2 + e^{2 x}

Add a constant to the solution

= ln 2 + e^{2 x} + C

x \to 0 lim (e^{9 x})

x \to - 1 lim ((\frac{2}{3})^{x})

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

\int_{\frac{1}{2}}^{3} 4 x ln (2 x) d x

\frac{x ^{2} - 4}{x}

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