integral from-3 to 0 of 1/(9+2x)

Lösung

\int_{- 3}^{0} \frac{1}{9 + 2 x} d x

\frac{1}{2} ln (3)

Dezimale

0.54930 \dots

Schritte zur Lösung

\int_{- 3}^{0} \frac{1}{9 + 2 x} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{3}^{9}

Berechne die Grenzen:

ln (3)

= \frac{1}{2} ln (3)

\int x e^{x} - x d x

\int arcsin (x + 2) d x

in v erse l a pl a ce \frac{1}{s ^{3} - 2 s ^{2} + 3 s}

p a r i t y y = arcsinh (tan (x))

t an g e n t f (x) = x cos (x), at x = \frac{3 π}{2}