integral from 0 to 1 of y/(1+xy)

Lösung

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

ln ∣ y + 1 ∣

Schritte zur Lösung

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

Entferne die Konstante:

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

= y \cdot \int_{0}^{1} \frac{1}{1 + y x} d x

Wende U-Substitution an

= y \cdot \int_{1}^{1 + y} \frac{1}{y u} d u

Entferne die Konstante:

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

= y \frac{1}{y} \cdot \int_{1}^{1 + y} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= y \frac{1}{y} [ln ∣ u ∣]_{1}^{1 + y}

Vereinfache

y \frac{1}{y} [ln ∣ u ∣]_{1}^{1 + y} : [ln ∣ u ∣]_{1}^{1 + y}

= [ln ∣ u ∣]_{1}^{1 + y}

Berechne die Grenzen:

ln ∣ y + 1 ∣

= ln ∣ y + 1 ∣

\int_{1}^{4} \frac{x}{1 + x} d x

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