integral from 0 to 1 of 1/(x^2+2x+5)

Lösung

\int_{0}^{1} \frac{1}{x ^{2} + 2 x + 5} d x

\frac{1}{2} (\frac{π}{4} - arctan (\frac{1}{2}))

Dezimale

0.16087 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{1}{x ^{2} + 2 x + 5} d x

Vervollständige das Quadrat

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

= \int_{0}^{1} \frac{1}{( x + 1 ) ^{2} + 4} d x

Wende U-Substitution an

= \int_{1}^{2} \frac{1}{u ^{2} + 4} d u

Wende integrale Substitution an

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

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{\frac{1}{2}}^{1} \frac{1}{v ^{2} + 1} d v

Nutze das gemeinsame Integral :

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

= \frac{1}{2} [arctan (v)]_{\frac{1}{2}}^{1}

Berechne die Grenzen:

\frac{π}{4} - arctan (\frac{1}{2})

= \frac{1}{2} (\frac{π}{4} - arctan (\frac{1}{2}))

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