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

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

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

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

arctan (e^{2 x}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 ( 1 + u ^{2} )} d u

= 2 \cdot \int \frac{1}{2 ( 1 + u ^{2} )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 2 \cdot \frac{1}{2} arctan (u)

Substitute back

u = e^{2 x}

= 2 \cdot \frac{1}{2} arctan (e^{2 x})

Simplify

2 \cdot \frac{1}{2} arctan (e^{2 x}) : arctan (e^{2 x})

= arctan (e^{2 x})

Add a constant to the solution

= arctan (e^{2 x}) + C

\frac{d y}{d x} = 5 - 8 x

\int \frac{a ^{x}}{b ^{x}} d x

\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}}

y^{2} = 2 x - 2, y = x - 5

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