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

Lösung

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

ln (3)

Dezimale

1.09861 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{3}

Berechne die Grenzen:

ln (3)

= ln (3)

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

\frac{d ^{4} y}{d x ^{4}} - 23 \frac{d ^{2} y}{d x ^{2}} - 50 y = 0

x \to 0 lim (sin (2 x))

in v erse l a pl a ce \frac{s}{( s - 2 ) ^{2} + 16}

\int_{\frac{1}{3}}^{5} 16 x ln (3 x) d x