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

Lösung

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

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

Schritte zur Lösung

\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

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 ^{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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{x ^{3} csc ^{2} ( x ^{4} )}{cot ( x ^{4} )} d x

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