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

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

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

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

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

Solution steps

\int \frac{e ^{3 x}}{e ^{6 x} + 1} 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 (e^{6 x} + 1 - 1)

Simplify

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

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