integral from 1 to 9 of 1/(xsqrt(x^2+9))

Lösung

\int_{1}^{9} \frac{1}{x x ^{2} + 9} d x

\frac{1}{3} (ln (tan (\frac{1}{2} arctan (3))) - ln (tan (\frac{1}{2} arctan (\frac{1}{3}))))

Dezimale

0.49699 \dots

Schritte zur Lösung

\int_{1}^{9} \frac{1}{x x ^{2} + 9} d x

Trigonometrische Substitution anwenden

= \int_{a r c t a n (\frac{1}{3})}^{a r c t a n (3)} \frac{sec ( u )}{3 tan ( u )} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{a r c t a n (\frac{1}{3})}^{a r c t a n (3)} \frac{sec ( u )}{tan ( u )} d u

Wende U-Substitution an

= \frac{1}{3} \cdot \int_{t a n (\frac{a r c t a n ( \frac{1}{3} )}{2})}^{t a n (\frac{a r c t a n ( 3 )}{2})} \frac{1}{v} d v

Nutze das gemeinsame Integral :

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} [ln ∣ v ∣]_{t a n (\frac{a r c t a n ( \frac{1}{3} )}{2})}^{t a n (\frac{a r c t a n ( 3 )}{2})}

Berechne die Grenzen:

ln (tan (\frac{1}{2} arctan (3))) - ln (tan (\frac{1}{2} arctan (\frac{1}{3})))

= \frac{1}{3} (ln (tan (\frac{1}{2} arctan (3))) - ln (tan (\frac{1}{2} arctan (\frac{1}{3}))))

t a y l or \frac{1}{1 + 2 x}

x \to 3 lim ((x - 0) \cdot (x - 3))

t a y l or sin (x), \frac{π}{6}

\int (e^{y^{3} + 2}) d y

\int_{0}^{4} (x) d x