What is the integral of (e^{4x})/(e^{4x)+6} ?

The integral of (e^{4x})/(e^{4x)+6} is 1/4 ln|e^{4x}+6|+C

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integral of (e^{4x})/(e^{4x)+6}

\int \frac{e ^{4 x}}{e ^{4 x} + 6} d x

\frac{1}{4} ln e^{4 x} + 6 + C

Solution steps

\int \frac{e ^{4 x}}{e ^{4 x} + 6} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = e^{4 x} + 6

= \frac{1}{4} ln e^{4 x} + 6

Add a constant to the solution

= \frac{1}{4} ln e^{4 x} + 6 + C

n = 1 \sum \infty \frac{5}{2 ^{n}}

y = \frac{2 x}{x + 1}, at x = 1

2 \frac{d y}{d x} + 4 y = 1

f (x) = cos (x^{4})

y = t

What is the integral of (e^{4x})/(e^{4x)+6} ?
The integral of (e^{4x})/(e^{4x)+6} is 1/4 ln|e^{4x}+6|+C