integral from 0 to x of (1+x)/(1-x)

Lösung

\int_{0}^{x} \frac{1 + x}{1 - x} d x

- 2 ln ∣ - x + 1 ∣ - x

Schritte zur Lösung

\int_{0}^{x} \frac{1 + x}{1 - x} d x

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

= \int_{0}^{x} \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_{0}^{x} \frac{2}{1 - x} d x - \int_{0}^{x} 1 d x

\int_{0}^{x} \frac{2}{1 - x} d x = - 2 ln ∣ - x + 1 ∣

\int_{0}^{x} 1 d x = x

= - 2 ln ∣ - x + 1 ∣ - x

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