integral from 0 to 1 of (6x)/(3x^2+1)

Lösung

\int_{0}^{1} \frac{6 x}{3 x ^{2} + 1} d x

2 ln (2)

Dezimale

1.38629 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{6 x}{3 x ^{2} + 1} d x

Entferne die Konstante:

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

= 6 \cdot \int_{0}^{1} \frac{x}{3 x ^{2} + 1} d x

Wende U-Substitution an

= 6 \cdot \int_{1}^{4} \frac{1}{6 u} d u

Entferne die Konstante:

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

= 6 \cdot \frac{1}{6} \cdot \int_{1}^{4} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} [ln ∣ u ∣]_{1}^{4}

Vereinfache

6 \cdot \frac{1}{6} [ln ∣ u ∣]_{1}^{4} : [ln ∣ u ∣]_{1}^{4}

= [ln ∣ u ∣]_{1}^{4}

Berechne die Grenzen:

2 ln (2)

= 2 ln (2)

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