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

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

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

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

\frac{1}{3} arctan (e^{3 x}) + C

Solution steps

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

Apply u-substitution

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

= \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

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

Apply u-substitution

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

= \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

= \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)

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

Substitute back

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

Simplify

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

= \frac{1}{3} arctan (e^{3 x})

Add a constant to the solution

= \frac{1}{3} arctan (e^{3 x}) + C

\frac{\partial}{\partial y} (y sin (x + y))

\frac{d}{d x} (e^{6 x} ln (x^{2} - 4))

\int_{- 1}^{2} (x - 6 ∣ x ∣) d x

\int \frac{1}{- x ^{2} + 6 x - 5} d x

4 x^{3} y + x^{4} y^{^{'}} = sin^{3} (\frac{x}{9})

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