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

Lösung

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

ln ∣ y + 2 ∣ - ln ∣ y + 1 ∣

Schritte zur Lösung

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{y + 1}^{y + 2}

Berechne die Grenzen:

ln ∣ y + 2 ∣ - ln ∣ y + 1 ∣

= ln ∣ y + 2 ∣ - ln ∣ y + 1 ∣

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

\int_{8}^{10} 0.5 x^{- 1.5} d x

\int_{- 2}^{2} (1 + 2 x) d x

\int_{1}^{2} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x

\int_{2}^{3} (3 t + 2) d t