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

Lösung

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

- x - ln ∣ - x + 1 ∣

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{1 - x} - \frac{- u + 1}{u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{1}^{1 - x} \frac{- u + 1}{u} d u

Multipliziere aus

\frac{- u + 1}{u} : - 1 + \frac{1}{u}

= - \int_{1}^{1 - x} - 1 + \frac{1}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int_{1}^{1 - x} 1 d u + \int_{1}^{1 - x} \frac{1}{u} d u)

\int_{1}^{1 - x} 1 d u = - x

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

= - (- (- x) + ln ∣ - x + 1 ∣)

Vereinfache

= - x - ln ∣ - x + 1 ∣

\int_{- \infty}^{0} e^{10 x} d x

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