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

Lösung

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

\frac{1}{4} ln (5)

Dezimale

0.40235 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{5} \frac{1}{4 u} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{1}^{5} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{1}^{5}

Berechne die Grenzen:

ln (5)

= \frac{1}{4} ln (5)

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