What is the integral of (e^{4x})/(3+e^{4x)} ?

The integral of (e^{4x})/(3+e^{4x)} is 1/4 ln|3+e^{4x}|+C

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integral of (e^{4x})/(3+e^{4x)}

\int \frac{e ^{4 x}}{3 + e ^{4 x}} d x

\frac{1}{4} ln 3 + e^{4 x} + C

Solution steps

\int \frac{e ^{4 x}}{3 + e ^{4 x}} 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 = 3 + e^{4 x}

= \frac{1}{4} ln 3 + e^{4 x}

Add a constant to the solution

= \frac{1}{4} ln 3 + e^{4 x} + C

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What is the integral of (e^{4x})/(3+e^{4x)} ?
The integral of (e^{4x})/(3+e^{4x)} is 1/4 ln|3+e^{4x}|+C