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

Lösung

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

divergiert

Schritte zur Lösung

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

Wende integrale Substitution an

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

Faktorisiere

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

= \frac{1}{2} \cdot \int_{- \frac{1}{2}}^{1} \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}}^{1} \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}}^{1})

Vereinfache

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

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

Berechne die Grenzen:

divergiert

= divergiert

\int_{1}^{x^{4}} sec (x) d x

\int_{0}^{x} \frac{1}{1 - t ^{2}} d t

\int_{1}^{x} \frac{10}{9 x ^{2}} d x

\int_{x - 3}^{x + 3} y d y

\int_{0}^{3} 53 - (3 x^{2} + 2 x + 20) d x