integral from 1 to 8 of 1/(xsqrt(16x^2-9))

Lösung

\int_{1}^{8} \frac{1}{x 16 x ^{2} - 9} d x

\frac{1}{3} (arcsec (\frac{32}{3}) - arcsec (\frac{4}{3}))

Dezimale

0.25139 \dots

Schritte zur Lösung

\int_{1}^{8} \frac{1}{x 16 x ^{2} - 9} d x

Trigonometrische Substitution anwenden

= \int_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})} \frac{1}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})} 1 d u

Integral einer Konstanten :

\int a d x = a x

= \frac{1}{3} [1 \cdot u]_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})}

Vereinfache

= \frac{1}{3} [u]_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})}

Berechne die Grenzen:

arcsec (\frac{32}{3}) - arcsec (\frac{4}{3})

= \frac{1}{3} (arcsec (\frac{32}{3}) - arcsec (\frac{4}{3}))

x \to 6 lim (cos (\frac{1}{x - 6}))

\int cos (x) sec^{3} (x) d x

d er i v a t i v e f (x) = \frac{x ^{3} + 3 x + 2}{x ^{2} - 1}

\int_{0}^{2} \int_{0}^{2 - x^{2}} 18 x y d y d x

x \to - 2 lim (\frac{x + 2}{2 x + x ^{2} + 12})