integral from 0 to 4 of (2x)/(x^2+1)

Lösung

\int_{0}^{4} \frac{2 x}{x ^{2} + 1} d x

ln (17)

Dezimale

2.83321 \dots

Schritte zur Lösung

\int_{0}^{4} \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}^{4} \frac{x}{x ^{2} + 1} d x

Wende U-Substitution an

= 2 \cdot \int_{1}^{17} \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}^{17} \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}^{17}

Vereinfache

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{17} : [ln ∣ u ∣]_{1}^{17}

= [ln ∣ u ∣]_{1}^{17}

Berechne die Grenzen:

ln (17)

= ln (17)

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