What is the integral of (e^{8x})/(e^{16x)+16} ?

The integral of (e^{8x})/(e^{16x)+16} is 1/32 arctan(1/4 e^{8x})+C

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integral of (e^{8x})/(e^{16x)+16}

\int \frac{e ^{8 x}}{e ^{16 x} + 16} d x

\frac{1}{32} arctan (\frac{1}{4} e^{8 x}) + C

Solution steps

\int \frac{e ^{8 x}}{e ^{16 x} + 16} d x

Apply u-substitution

: \int \frac{e ^{u}}{8 ( e ^{2 u} + 16 )} d u

= \int \frac{e ^{u}}{8 ( e ^{2 u} + 16 )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{e ^{u}}{e ^{2 u} + 16} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 16} d v

= \frac{1}{8} \cdot \int \frac{1}{v ^{2} + 16} d v

Apply Integral Substitution

= \frac{1}{8} \cdot \int \frac{1}{4 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{4} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{8} \cdot \frac{1}{4} arctan (w)

Substitute back

= \frac{1}{8} \cdot \frac{1}{4} arctan (\frac{e ^{8 x}}{4})

Simplify

\frac{1}{8} \cdot \frac{1}{4} arctan (\frac{e ^{8 x}}{4}) : \frac{1}{32} arctan (\frac{1}{4} e^{8 x})

= \frac{1}{32} arctan (\frac{1}{4} e^{8 x})

Add a constant to the solution

= \frac{1}{32} arctan (\frac{1}{4} e^{8 x}) + C

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What is the integral of (e^{8x})/(e^{16x)+16} ?
The integral of (e^{8x})/(e^{16x)+16} is 1/32 arctan(1/4 e^{8x})+C