integral of 1/(1+e^{9x)}

Lösung

\int \frac{1}{1 + e ^{9 x}} d x

- \frac{1}{9} ln 1 + e^{9 x} + x + C

Schritte zur Lösung

\int \frac{1}{1 + e ^{9 x}} d x

\frac{1}{1 + e ^{9 x}} = - \frac{e ^{9 x}}{1 + e ^{9 x}} + 1

= \int - \frac{e ^{9 x}}{1 + e ^{9 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 ^{9 x}}{1 + e ^{9 x}} d x + \int 1 d x

\int \frac{e ^{9 x}}{1 + e ^{9 x}} d x = \frac{1}{9} ln 1 + e^{9 x}

\int 1 d x = x

= - \frac{1}{9} ln 1 + e^{9 x} + x

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{9} ln 1 + e^{9 x} + x + C

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