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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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