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

Lösung

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

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

Dezimale

- 0.54930 \dots

Schritte zur Lösung

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

Wende integrale Substitution an

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

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}}^{\frac{1}{2}} \frac{1}{u ^{2} - 1} d u

Faktorisiere

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

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

Vereinfache

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

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

Berechne die Grenzen:

\frac{2 ln ( 2 ) + 2 ln ( \frac{3}{2} )}{2}

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

Vereinfache

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

\int_{1}^{e^{2}} \frac{( ln ( x ) ) ^{2}}{x} d x

\int_{0}^{1} sin (π) x d x

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