integral of (e^{2x}+2e^x)/(e^{2x)-8}

Lösung

\int \frac{e ^{2 x} + 2 e ^{x}}{e ^{2 x} - 8} d x

\frac{1}{2} ln e^{2 x} - 8 - \frac{1}{2 2} (ln \frac{e ^{x}}{2 2} + 1 - ln \frac{e ^{x}}{2 2} - 1) + C

Schritte zur Lösung

\int \frac{e ^{2 x} + 2 e ^{x}}{e ^{2 x} - 8} d x

Multipliziere aus

\frac{e ^{2 x} + 2 e ^{x}}{e ^{2 x} - 8} : \frac{e ^{2 x}}{e ^{2 x} - 8} + \frac{2 e ^{x}}{e ^{2 x} - 8}

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}}{e ^{2 x} - 8} d x + \int \frac{2 e ^{x}}{e ^{2 x} - 8} d x

\int \frac{e ^{2 x}}{e ^{2 x} - 8} d x = \frac{1}{2} ln e^{2 x} - 8

\int \frac{2 e ^{x}}{e ^{2 x} - 8} d x = - \frac{1}{2 2} (ln \frac{e ^{x}}{2 2} + 1 - ln \frac{e ^{x}}{2 2} - 1)

= \frac{1}{2} ln e^{2 x} - 8 - \frac{1}{2 2} (ln \frac{e ^{x}}{2 2} + 1 - ln \frac{e ^{x}}{2 2} - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln e^{2 x} - 8 - \frac{1}{2 2} (ln \frac{e ^{x}}{2 2} + 1 - ln \frac{e ^{x}}{2 2} - 1) + C

y^{^{'}} - y = 5 x - 5, y (0) = 9

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

\int_{- 3}^{3} x (9 - x^{2}) d x

\int 5 x^{3} (x - 1)^{- 4} d x

n = 1 \sum \infty \frac{1}{n ^{6}}