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

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

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

Solution steps

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

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

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

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

\int 1 d x = x

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

Add a constant to the solution

= arctan (e^{x}) - \frac{1}{2} ln 1 + e^{2 x} + 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 arctan(e^x)-1/2 ln|1+e^{2x}|+x+C