integral from 2 to 5 of 1/(x+1)

Lösung

\int_{2}^{5} \frac{1}{x + 1} d x

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

\int_{2}^{5} \frac{1}{x + 1} d x

Wende U-Substitution an

= \int_{3}^{6} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{6}

Berechne die Grenzen:

ln (6) - ln (3)

= ln (6) - ln (3)

Vereinfache

= ln (2)

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 2 y ^{2}})

\int \frac{81 - x ^{2}}{x ^{2}} d x

x \to - 1 lim (x)

d er i v a t i v e f (x) = \frac{5 x ^{\frac{3}{2}} - 1 5 ^{\frac{1}{2}}}{2}

\int_{- 5}^{5} 25 - x^{2} d x