integral from 0 to 8 of (2x)/(x^2+1)

Lösung

\int_{0}^{8} \frac{2 x}{x ^{2} + 1} d x

ln (65)

Dezimale

4.17438 \dots

Schritte zur Lösung

\int_{0}^{8} \frac{2 x}{x ^{2} + 1} d x

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{8} \frac{x}{x ^{2} + 1} d x

Wende U-Substitution an

= 2 \cdot \int_{1}^{65} \frac{1}{2 u} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{2} \cdot \int_{1}^{65} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{65}

Vereinfache

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{65} : [ln ∣ u ∣]_{1}^{65}

= [ln ∣ u ∣]_{1}^{65}

Berechne die Grenzen:

ln (65)

= ln (65)

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