What is the integral of (3e^{2x})/(1+e^{2x)} ?

The integral of (3e^{2x})/(1+e^{2x)} is 3/2 ln|1+e^{2x}|+C

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integral of (3e^{2x})/(1+e^{2x)}

\int \frac{3 e ^{2 x}}{1 + e ^{2 x}} d x

\frac{3}{2} ln 1 + e^{2 x} + C

Solution steps

\int \frac{3 e ^{2 x}}{1 + e ^{2 x}} d x

Take the constant out:

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

= 3 \cdot \int \frac{e ^{2 x}}{1 + e ^{2 x}} d x

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + e^{2 x}

= 3 \cdot \frac{1}{2} ln 1 + e^{2 x}

Simplify

3 \cdot \frac{1}{2} ln 1 + e^{2 x} : \frac{3}{2} ln 1 + e^{2 x}

= \frac{3}{2} ln 1 + e^{2 x}

Add a constant to the solution

= \frac{3}{2} ln 1 + e^{2 x} + C

f (x) = sin (2 x), 0, \frac{π}{2}

\frac{d}{d x} (5 sin (x) + 3 cos (x))

\frac{d}{d x} (e^{- x y})

ln (4 + x)

x \to 0 lim (x (e^{\frac{1}{x}} - 1))

What is the integral of (3e^{2x})/(1+e^{2x)} ?
The integral of (3e^{2x})/(1+e^{2x)} is 3/2 ln|1+e^{2x}|+C