What is the integral of (e^{4x})/(e^{4x)+1} ?

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

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integral of (e^{4x})/(e^{4x)+1}

\int \frac{e ^{4 x}}{e ^{4 x} + 1} d x

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

Solution steps

\int \frac{e ^{4 x}}{e ^{4 x} + 1} 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} + 1

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

Add a constant to the solution

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

\frac{d}{d x} (7 e^{- 3 x})

\int x^{- \frac{5}{4}} d x

\int cos^{6} (x) d x

(x + \frac{y}{x}) d x - d y = 0

y^{^{'}} = 2 y + e^{2 x} cos (3 x), y (0) = 4

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