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

Lösung

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

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

Schritte zur Lösung

\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

Wende die Summenregel an:

\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

Füge eine Konstante zur Lösung hinzu

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

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