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

The integral of (e^2x)/(4+e^4x) is 1/(e^6)(4+e^4x-4ln|4+e^4x|)+C

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integral of (e^2x)/(4+e^4x)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 4}{e ^{8} u} d u

= e^{2} \cdot \int \frac{u - 4}{e ^{8} u} d u

Take the constant out:

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

= e^{2} \frac{1}{e ^{8}} \cdot \int \frac{u - 4}{u} d u

Expand

\frac{u - 4}{u} : 1 - \frac{4}{u}

= e^{2} \frac{1}{e ^{8}} \cdot \int 1 - \frac{4}{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

= e^{2} \frac{1}{e ^{8}} (\int 1 d u - \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= e^{2} \frac{1}{e ^{8}} (u - 4 ln ∣ u ∣)

Substitute back

u = 4 + e^{4} x

= e^{2} \frac{1}{e ^{8}} (4 + e^{4} x - 4 ln 4 + e^{4} x)

Simplify

e^{2} \frac{1}{e ^{8}} (4 + e^{4} x - 4 ln 4 + e^{4} x) : \frac{1}{e ^{6}} (4 + e^{4} x - 4 ln 4 + e^{4} x)

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{( 2 x + 1 )}{2 y}

9 x (2 x^{2} + 1)

\int (x ln (x))^{2} (\frac{1}{x ^{2}}) d x

2 x

\int 9 t^{2} + 4 d t

What is the integral of (e^2x)/(4+e^4x) ?
The integral of (e^2x)/(4+e^4x) is 1/(e^6)(4+e^4x-4ln|4+e^4x|)+C