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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + e^{2 x}

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

Add a constant to the solution

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

f (x) = 4 x^{2} - 4 x + 1, at x = 1

\frac{\partial}{\partial x} (x sin (x + y) - cos (x + y))

y^{^{'}} = y^{4} \cdot (4 x^{2} + x)

(7, 4), (1, - 2)

\int (x^{2} - 2 x)^{2} d x

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