integral from 1 to 7 of 1/(x+1)

Lösung

\int_{1}^{7} \frac{1}{x + 1} d x

2 ln (2)

Dezimale

1.38629 \dots

Schritte zur Lösung

\int_{1}^{7} \frac{1}{x + 1} d x

Wende U-Substitution an

= \int_{2}^{8} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{8}

Berechne die Grenzen:

2 ln (2)

= 2 ln (2)

\int_{- 1}^{1} x^{2021} d x

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

\int_{0}^{l n (A)} 6 e^{2 x} d x

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

\int_{0}^{1} t - 1 d t