integral of 2/(e^{2x)+6}+1/(e^{2x)-3}

Lösung

\int \frac{2}{e ^{2 x} + 6} + \frac{1}{e ^{2 x} - 3} d x

- \frac{1}{6} ln e^{2 x} + 6 + \frac{x}{3} + \frac{1}{6} (ln e^{2 x} - 3 - 2 x) + C

Schritte zur Lösung

\int \frac{2}{e ^{2 x} + 6} + \frac{1}{e ^{2 x} - 3} 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{2}{e ^{2 x} + 6} d x + \int \frac{1}{e ^{2 x} - 3} d x

\int \frac{2}{e ^{2 x} + 6} d x = - \frac{1}{6} ln e^{2 x} + 6 + \frac{x}{3}

\int \frac{1}{e ^{2 x} - 3} d x = \frac{1}{6} (ln e^{2 x} - 3 - 2 x)

= - \frac{1}{6} ln e^{2 x} + 6 + \frac{x}{3} + \frac{1}{6} (ln e^{2 x} - 3 - 2 x)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{6} ln e^{2 x} + 6 + \frac{x}{3} + \frac{1}{6} (ln e^{2 x} - 3 - 2 x) + C

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