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

Lösung

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

- \frac{ln ( 10 )}{6}

Dezimale

- 0.38376 \dots

Schritte zur Lösung

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

Wende integrale Substitution an

= \int_{- \frac{1}{3}}^{\frac{2}{3}} \frac{1}{3 ( u ^{2} - 1 )} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{- \frac{1}{3}}^{\frac{2}{3}} \frac{1}{u ^{2} - 1} d u

Faktorisiere

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{3} \cdot \int_{- \frac{1}{3}}^{\frac{2}{3}} \frac{1}{- ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{3} (- \int_{- \frac{1}{3}}^{\frac{2}{3}} \frac{1}{- u ^{2} + 1} d u)

Nutze das gemeinsame Integral :

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{3} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{1}{3}}^{\frac{2}{3}})

Vereinfache

\frac{1}{3} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{1}{3}}^{\frac{2}{3}}) : - \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{1}{3}}^{\frac{2}{3}}

= - \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{1}{3}}^{\frac{2}{3}}

Berechne die Grenzen:

\frac{ln ( \frac{5}{3} ) + ln ( 3 ) - ln ( \frac{2}{3} ) + ln ( \frac{4}{3} )}{2}

= - \frac{1}{3} \cdot \frac{ln ( \frac{5}{3} ) + ln ( 3 ) - ln ( \frac{2}{3} ) + ln ( \frac{4}{3} )}{2}

Vereinfache

= - \frac{ln ( 10 )}{6}

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

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\int_{0}^{0.5} \frac{1}{x ln ^{2} ∣ x ∣} d x

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