integral of 1/(2+e^x)

Lösung

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

\frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣ + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Ermittle den Partialbruch von

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

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

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= \frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣

Setze in

u = e^{x}

ein

= \frac{1}{2} ln ∣ e^{x} ∣ - \frac{1}{2} ln ∣ e^{x} + 2 ∣

\frac{1}{2} ln ∣ e^{x} ∣ = \frac{1}{2} x

= \frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} x - \frac{1}{2} ln ∣ e^{x} + 2 ∣ + C

\int t + 6 d t

d er i v a t i v e f (x) = \frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}}

x \to 0 lim (cot^{3} (x))

\frac{d}{d x} (\frac{tan ( 20 x )}{x ^{\frac{2}{3}} + 3 x})

x \to 0 lim (\frac{x}{ln ( x + 1 )})