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

Lösung

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

- 2 ln ∣ 1 - x ∣ - x + C

Schritte zur Lösung

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

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

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

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

\int 1 d x = x

= - 2 ln ∣ 1 - x ∣ - x

Füge eine Konstante zur Lösung hinzu

= - 2 ln ∣ 1 - x ∣ - x + C

d er i v a t i v e f (x) = (x - 1)^{2} - 2

\frac{\partial}{\partial x} (2 y e^{7 x})

x \to 0 lim (tan^{x} (6 x))

\int (3 x)^{5} (3 x)^{3} + 1 d x

d er i v a t i v e f (x) = \frac{98}{x ^{3}}