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

Lösung

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

\frac{1}{2} (ln (7) - ln (3))

Dezimale

0.42364 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{3}^{7} \frac{1}{2 u} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{3}^{7} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{3}^{7}

Berechne die Grenzen:

ln (7) - ln (3)

= \frac{1}{2} (ln (7) - ln (3))

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