integral from-2 to e-3 of 1/(t+3)

Lösung

\int_{- 2}^{e - 3} \frac{1}{t + 3} d t

1

Schritte zur Lösung

\int_{- 2}^{e - 3} \frac{1}{t + 3} d t

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e}

Berechne die Grenzen:

1

= 1

\int_{- 2}^{2} x cos (\frac{nπ x}{2}) d x

\int_{1}^{e} x^{- 1} d x

\int_{0}^{π} 6 e^{x} + 7 sin (x) d x

\int_{3}^{\infty} e^{- s t} 2 d t

\int_{- \infty}^{- 1} \frac{x}{x ^{4} + 1} d x