What is the integral from 0 to ln(2) of (e^{3x})/(1+e^{6x)} ?

The integral from 0 to ln(2) of (e^{3x})/(1+e^{6x)} is 1/3 (arctan(8)-pi/4)

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integral from 0 to ln(2) of (e^{3x})/(1+e^{6x)}

\int_{0}^{l n (2)} \frac{e ^{3 x}}{1 + e ^{6 x}} d x

\frac{1}{3} (arctan (8) - \frac{π}{4})

Decimal Notation

0.22034 \dots

Solution steps

\int_{0}^{l n (2)} \frac{e ^{3 x}}{1 + e ^{6 x}} d x

Apply u-substitution

: \int_{2}^{65} \frac{1}{6 u u - 1} d u

= \int_{2}^{65} \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_{2}^{65} \frac{1}{u u - 1} d u

Apply u-substitution

: \int_{1}^{8} \frac{2}{v ^{2} + 1} d v

= \frac{1}{6} \cdot \int_{1}^{8} \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_{1}^{8} \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)]_{1}^{8}

Simplify

\frac{1}{6} \cdot 2 [arctan (v)]_{1}^{8} : \frac{1}{3} [arctan (v)]_{1}^{8}

= \frac{1}{3} [arctan (v)]_{1}^{8}

Compute the boundaries:

arctan (8) - \frac{π}{4}

= \frac{1}{3} (arctan (8) - \frac{π}{4})

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\frac{\partial}{\partial x} (- 36 y sin (9 x))

\frac{\partial}{\partial x} (3 x - 2 y^{4})

What is the integral from 0 to ln(2) of (e^{3x})/(1+e^{6x)} ?
The integral from 0 to ln(2) of (e^{3x})/(1+e^{6x)} is 1/3 (arctan(8)-pi/4)