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

Lösung

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

\frac{1}{37} ln ∣ e^{x} - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln e^{2 x} + 1 - 6 arctan (e^{x})) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{1}{( u - 6 ) ( u ^{2} + 1 )} d u

Ermittle den Partialbruch von

\frac{1}{( u - 6 ) ( u ^{2} + 1 )} : \frac{1}{37 ( u - 6 )} + \frac{- u - 6}{37 ( u ^{2} + 1 )}

= \int \frac{1}{37 ( u - 6 )} + \frac{- u - 6}{37 ( 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}{37 ( u - 6 )} d u + \int \frac{- u - 6}{37 ( u ^{2} + 1 )} d u

\int \frac{1}{37 ( u - 6 )} d u = \frac{1}{37} ln ∣ u - 6 ∣

\int \frac{- u - 6}{37 ( u ^{2} + 1 )} d u = \frac{1}{37} (- \frac{1}{2} ln u^{2} + 1 - 6 arctan (u))

= \frac{1}{37} ln ∣ u - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln u^{2} + 1 - 6 arctan (u))

Setze in

u = e^{x}

ein

= \frac{1}{37} ln ∣ e^{x} - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln (e^{x})^{2} + 1 - 6 arctan (e^{x}))

Wende Exponentenregel an:

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

= \frac{1}{37} ln ∣ e^{x} - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln e^{2 x} + 1 - 6 arctan (e^{x}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{37} ln ∣ e^{x} - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln e^{2 x} + 1 - 6 arctan (e^{x})) + C

d er i v a t i v e f (x) = \frac{x}{x + 17}

\int 3 x - 2 x d x

\frac{\partial}{\partial y} (y e^{x y})

x \to 0 lim (sin (1))

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