integral of ((1+e^x))/((1-e^x))

Lösung

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

- 2 ln ∣ 1 - e^{x} ∣ + x + C

Schritte zur Lösung

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

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

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

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

\int 1 d x = x

= - 2 ln ∣ 1 - e^{x} ∣ + x

Füge eine Konstante zur Lösung hinzu

= - 2 ln ∣ 1 - e^{x} ∣ + x + C

\int \frac{cos ( u )}{sin ( u ) + sin ^{3} ( u )} d u

\int e^{\frac{- x ^{2}}{5}} d x

\int 12 t e^{t} d t

\int \frac{2 x - 5}{x ^{2} + x + 1} d x

\int \frac{4 x}{2 x ^{2}} d x