integral from 2 to 4 of 3/(2x+1)

Lösung

\int_{2}^{4} \frac{3}{2 x + 1} d x

3 ln (3) - \frac{3}{2} ln (5)

Dezimale

0.88167 \dots

Schritte zur Lösung

\int_{2}^{4} \frac{3}{2 x + 1} d x

Entferne die Konstante:

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

= 3 \cdot \int_{2}^{4} \frac{1}{2 x + 1} d x

Wende U-Substitution an

= 3 \cdot \int_{5}^{9} \frac{1}{2 u} d u

Entferne die Konstante:

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

= 3 \cdot \frac{1}{2} \cdot \int_{5}^{9} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{9}

Vereinfache

3 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{9} : \frac{3}{2} [ln ∣ u ∣]_{5}^{9}

= \frac{3}{2} [ln ∣ u ∣]_{5}^{9}

Berechne die Grenzen:

2 ln (3) - ln (5)

= \frac{3}{2} (2 ln (3) - ln (5))

Vereinfache

= 3 ln (3) - \frac{3}{2} ln (5)

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