integral from 1 to 6 of 1/(sqrt(1+x^2))

Lösung

\int_{1}^{6} \frac{1}{1 + x ^{2}} d x

ln (37 + 6) - ln (2 + 1)

Dezimale

1.61040 \dots

Schritte zur Lösung

\int_{1}^{6} \frac{1}{1 + x ^{2}} d x

Nutze das gemeinsame Integral :

\int \frac{1}{1 + x ^{2}} d x = ln x^{2} + 1 + x

= [ln x^{2} + 1 + x]_{1}^{6}

Berechne die Grenzen:

ln (37 + 6) - ln (2 + 1)

= ln (37 + 6) - ln (2 + 1)

\int_{- 1}^{2} 2 x^{2} - 4 x + 5 d x

\int_{0}^{8} \frac{1}{v ^{\frac{1}{3}} + 1} d v

\int_{0}^{2} π (x^{2} - 4)^{2} d x

\int_{0}^{2 π} 9 d x

\int_{0}^{1} (x^{\frac{1}{2}}) d x