integral of (e^x)/((e^x-9)(e^{2x)+1)}

Lösung

\int \frac{e ^{x}}{( e ^{x} - 9 ) ( e ^{2 x} + 1 )} d x

\frac{1}{82} ln ∣ e^{x} - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln e^{2 x} + 1 - 9 arctan (e^{x})) + C

Schritte zur Lösung

\int \frac{e ^{x}}{( e ^{x} - 9 ) ( e ^{2 x} + 1 )} d x

Wende U-Substitution an

= \int \frac{1}{( u - 9 ) ( u ^{2} + 1 )} d u

Ermittle den Partialbruch von

\frac{1}{( u - 9 ) ( u ^{2} + 1 )} : \frac{1}{82 ( u - 9 )} + \frac{- u - 9}{82 ( u ^{2} + 1 )}

= \int \frac{1}{82 ( u - 9 )} + \frac{- u - 9}{82 ( u ^{2} + 1 )} d u

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{1}{82 ( u - 9 )} d u + \int \frac{- u - 9}{82 ( u ^{2} + 1 )} d u

\int \frac{1}{82 ( u - 9 )} d u = \frac{1}{82} ln ∣ u - 9 ∣

\int \frac{- u - 9}{82 ( u ^{2} + 1 )} d u = \frac{1}{82} (- \frac{1}{2} ln u^{2} + 1 - 9 arctan (u))

= \frac{1}{82} ln ∣ u - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln u^{2} + 1 - 9 arctan (u))

Setze in

u = e^{x}

ein

= \frac{1}{82} ln ∣ e^{x} - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln (e^{x})^{2} + 1 - 9 arctan (e^{x}))

Wende Exponentenregel an:

(a^{b})^{c} = a^{b c}

= \frac{1}{82} ln ∣ e^{x} - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln e^{2 x} + 1 - 9 arctan (e^{x}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{82} ln ∣ e^{x} - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln e^{2 x} + 1 - 9 arctan (e^{x})) + C

\frac{d}{d x} (\frac{1}{6})

d er i v a t i v e (1 + 4 x^{2}) (x - x^{2})

\frac{d}{d x} (A cos (x) + B sin (x))

\int_{- 1}^{6} - x^{2} + 5 x + 6 d x

a re a y = \frac{1}{2} x, y = \frac{6}{x}