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

\int (e^{x} - 1)^{2} d x

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

d er i v a t i v e \frac{6}{x ^{5}} - \frac{2}{x}

\frac{\partial}{\partial x} (3 (x + 1)^{2})

\frac{\partial}{\partial x} (z^{3} - x^{2} y)