integral from 0 to 2 of 1/(x^2+6x+13)

Lösung

\int_{0}^{2} \frac{1}{x ^{2} + 6 x + 13} d x

\frac{1}{2} (arctan (\frac{5}{2}) - arctan (\frac{3}{2}))

Dezimale

0.10374 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{1}{x ^{2} + 6 x + 13} d x

Vervollständige das Quadrat

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= \int_{0}^{2} \frac{1}{( x + 3 ) ^{2} + 4} d x

Wende U-Substitution an

= \int_{3}^{5} \frac{1}{u ^{2} + 4} d u

Wende integrale Substitution an

= \int_{\frac{3}{2}}^{\frac{5}{2}} \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{3}{2}}^{\frac{5}{2}} \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{3}{2}}^{\frac{5}{2}}

Berechne die Grenzen:

arctan (\frac{5}{2}) - arctan (\frac{3}{2})

= \frac{1}{2} (arctan (\frac{5}{2}) - arctan (\frac{3}{2}))

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