What is the integral of (e^{4x})/(1+e^{2x)} ?

The integral of (e^{4x})/(1+e^{2x)} is 1/2 (1+e^{2x}-ln|1+e^{2x}|)+C

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integral of (e^{4x})/(1+e^{2x)}

\int \frac{e ^{4 x}}{1 + e ^{2 x}} d x

\frac{1}{2} (1 + e^{2 x} - ln 1 + e^{2 x}) + C

Solution steps

\int \frac{e ^{4 x}}{1 + e ^{2 x}} d x

Apply u-substitution

: \int \frac{u - 1}{2 u} d u

= \int \frac{u - 1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{2} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{2} (u - ln ∣ u ∣)

Substitute back

u = 1 + e^{2 x}

= \frac{1}{2} (1 + e^{2 x} - ln 1 + e^{2 x})

Add a constant to the solution

= \frac{1}{2} (1 + e^{2 x} - ln 1 + e^{2 x}) + C

\int_{0}^{\infty} \frac{x}{e ^{x}} d x

\int_{1}^{e} ln (4 x) d x

(1 - x^{2}) \frac{d y}{d x} = x e^{y}

x \to + 5 lim (\frac{( x + 1 )}{x - 5})

x = 2 3 y, x = 0, y = 9

What is the integral of (e^{4x})/(1+e^{2x)} ?
The integral of (e^{4x})/(1+e^{2x)} is 1/2 (1+e^{2x}-ln|1+e^{2x}|)+C