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

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

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

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

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

Solution steps

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

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

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

Apply the Sum Rule:

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\frac{1}{3} x, x + 4, [4, - 1]

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

y^{^{''}} + 6 y^{^{'}} - 7 y = 0, y (0) = 1, y^{^{'}} (0) = 1

f (x) = \frac{5}{4 x}

\frac{d}{d x} (\frac{2 x + 9}{9 - x})

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