integral from 0 to 1 of 1/(x^2-3)

Lösung

\int_{0}^{1} \frac{1}{x ^{2} - 3} d x

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

Dezimale

- 0.38017 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{1}{x ^{2} - 3} d x

Wende integrale Substitution an

= \int_{0}^{\frac{1}{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_{0}^{\frac{1}{3}} \frac{1}{u ^{2} - 1} d u

Faktorisiere

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

= \frac{1}{3} \cdot \int_{0}^{\frac{1}{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_{0}^{\frac{1}{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}]_{0}^{\frac{1}{3}})

Vereinfache

\frac{1}{3} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{3}}) : - \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{3}}

= - \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{3}}

Berechne die Grenzen:

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

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

Vereinfache

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

\int \frac{9 x ^{2} + x + 48}{( x + 3 ) ( x ^{2} + 9 )} d x

\frac{\partial}{\partial y} (x y e^{s i n (π x y)})

a re a y = 2, y = sec (x)

\frac{d}{d t} (sin^{2} (π t))

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} + z ^{2}})