integral of (2+e^x)/(e^x)

Lösung

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

- \frac{2}{e ^{x}} + x + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{2 + u}{u ^{2}} d u

Multipliziere aus

\frac{2 + u}{u ^{2}} : \frac{2}{u ^{2}} + \frac{1}{u}

= \int \frac{2}{u ^{2}} + \frac{1}{u} 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{2}{u ^{2}} d u + \int \frac{1}{u} d u

\int \frac{2}{u ^{2}} d u = - \frac{2}{u}

\int \frac{1}{u} d u = ln ∣ u ∣

= - \frac{2}{u} + ln ∣ u ∣

Setze in

u = e^{x}

ein

= - \frac{2}{e ^{x}} + ln ∣ e^{x} ∣

ln ∣ e^{x} ∣ = x

= - \frac{2}{e ^{x}} + x

Füge eine Konstante zur Lösung hinzu

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

n = 1 \sum \infty \frac{7}{2 ^{n - 1}}

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

x \to a lim ([f (x)]^{g (x)})

t \to 0 lim (\frac{3}{t} - \frac{3}{e ^{t} - 1})

y^{^{'}} + y = t e^{- t} + 1