integral from 4 to 6 of 1/(sqrt(x^2-9))

Lösung

\int_{4}^{6} \frac{1}{x ^{2} - 9} d x

ln (- \frac{( 2 + 3 ) ( 7 - 4 )}{3})

Dezimale

0.52159 \dots

Schritte zur Lösung

\int_{4}^{6} \frac{1}{x ^{2} - 9} d x

Trigonometrische Substitution anwenden

= \int_{arcsec (\frac{4}{3})}^{\frac{π}{3}} sec (u) d u

Nutze das gemeinsame Integral :

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= [ln ∣ tan (u) + sec (u) ∣]_{arcsec (\frac{4}{3})}^{\frac{π}{3}}

Berechne die Grenzen:

ln (3 + 2) - ln (\frac{7}{3} + \frac{4}{3})

= ln (3 + 2) - ln (\frac{7}{3} + \frac{4}{3})

Vereinfache

= ln (- \frac{( 2 + 3 ) ( 7 - 4 )}{3})

\int_{1}^{\infty} \frac{1}{x ^{0.99}} d x

\int_{0}^{4} (6 + 7 x) d x

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

\int_{0}^{1} 2 t d t

\int_{0}^{\frac{π}{3}} sin^{6} (x) d x