integral from-2 to-1 of 1/(x^2+4x+5)

Lösung

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

\frac{π}{4}

Dezimale

0.78539 \dots

Schritte zur Lösung

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

Vervollständige das Quadrat

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

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

Wende U-Substitution an

= \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{π}{4}

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