What is the integral of (2e^{3x})/(1+e^{6x)} ?

The integral of (2e^{3x})/(1+e^{6x)} is 2/3 arctan(e^{3x})+C

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integral of (2e^{3x})/(1+e^{6x)}

\int \frac{2 e ^{3 x}}{1 + e ^{6 x}} d x

\frac{2}{3} arctan (e^{3 x}) + C

Solution steps

\int \frac{2 e ^{3 x}}{1 + e ^{6 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 ^{3 x}}{1 + e ^{6 x}} d x

Apply u-substitution

: \int \frac{1}{6 u u - 1} d u

= 2 \cdot \int \frac{1}{6 u u - 1} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{u u - 1} d u

Apply u-substitution

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

= 2 \cdot \frac{1}{6} \cdot \int \frac{2}{v ^{2} + 1} d v

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot 2 \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot \frac{1}{6} \cdot 2 arctan (v)

Substitute back

= 2 \cdot \frac{1}{6} \cdot 2 arctan (1 + e^{6 x} - 1)

Simplify

2 \cdot \frac{1}{6} \cdot 2 arctan (1 + e^{6 x} - 1) : \frac{2}{3} arctan (e^{3 x})

= \frac{2}{3} arctan (e^{3 x})

Add a constant to the solution

= \frac{2}{3} arctan (e^{3 x}) + C

\frac{d}{d x} (\frac{0.5}{cos ( x )})

\frac{d y}{d x} = \frac{ln ( x )}{x y}

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

\int \frac{1}{20 y} d y

\frac{d}{d x} (7^{2 x - 1})

What is the integral of (2e^{3x})/(1+e^{6x)} ?
The integral of (2e^{3x})/(1+e^{6x)} is 2/3 arctan(e^{3x})+C