What is the integral of (e^{8x})/(e^{8x)+6} ?

The integral of (e^{8x})/(e^{8x)+6} is 1/8 ln|e^{8x}+6|+C

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integral of (e^{8x})/(e^{8x)+6}

\int \frac{e ^{8 x}}{e ^{8 x} + 6} d x

\frac{1}{8} ln e^{8 x} + 6 + C

Solution steps

\int \frac{e ^{8 x}}{e ^{8 x} + 6} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} 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{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = e^{8 x} + 6

= \frac{1}{8} ln e^{8 x} + 6

Add a constant to the solution

= \frac{1}{8} ln e^{8 x} + 6 + C

y \to 1 lim (\frac{2 y ^{2} + 3}{y - 1})

\frac{d}{d t} (2 e^{- t} t - e^{- t} t^{2})

f (x) = ln (2 x^{5} (5 x - 1)^{3})

y = 3 x + 1

\frac{d}{d x} ((9 x^{6} + 4 x^{3})^{4})

What is the integral of (e^{8x})/(e^{8x)+6} ?
The integral of (e^{8x})/(e^{8x)+6} is 1/8 ln|e^{8x}+6|+C