integral from 1 to 9 of 1/(x+sqrt(x))

Lösung

\int_{1}^{9} \frac{1}{x + x} d x

2 ln (2)

Dezimale

1.38629 \dots

Schritte zur Lösung

\int_{1}^{9} \frac{1}{x + x} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= 2 \cdot \int_{2}^{4} \frac{1}{v} d v

Nutze das gemeinsame Integral :

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 [ln ∣ v ∣]_{2}^{4}

Berechne die Grenzen:

ln (2)

= 2 ln (2)

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