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

Lösung

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

divergiert

Schritte zur Lösung

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

Faktorisiere

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

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

= - [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{1}^{4}

Vereinfache

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{1}^{4} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{1}^{4}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{1}^{4}

Berechne die Grenzen:

divergiert

= divergiert

\int_{- 1}^{3} 2 x + 6 d x

\int_{0}^{1} (x - x e^{x - 1}) d x

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

\int_{- 1}^{3} 2 x + 4 d x

\int_{0}^{3} (y - 3) (2 y + 3) d y