integral of 1/(2^x+1)

Lösung

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

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

Schritte zur Lösung

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

\frac{1}{2 ^{x} + 1} = - \frac{2 ^{x}}{2 ^{x} + 1} + 1

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

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

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

\int 1 d x = x

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

Füge eine Konstante zur Lösung hinzu

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

e x p an d (x + 19) e^{3 x} + 4

m \to - 1 lim (m^{3} - 1)

a re a f (x) = x^{2} - x^{3}, [- 2, 0]

\frac{d}{d x} (7 x - 3)

\int 9 x e^{2 x} d x