integral from 2 to 4 of 1/(x^2-25)

Lösung

\int_{2}^{4} \frac{1}{x ^{2} - 25} d x

- \frac{2 ln ( 3 ) - ln ( \frac{7}{5} ) + ln ( \frac{3}{5} )}{10}

Dezimale

- 0.13499 \dots

Schritte zur Lösung

\int_{2}^{4} \frac{1}{x ^{2} - 25} d x

Wende integrale Substitution an

= \int_{\frac{2}{5}}^{\frac{4}{5}} \frac{1}{5 ( u ^{2} - 1 )} d u

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int_{\frac{2}{5}}^{\frac{4}{5}} \frac{1}{u ^{2} - 1} d u

Faktorisiere

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

= \frac{1}{5} \cdot \int_{\frac{2}{5}}^{\frac{4}{5}} \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}{5} (- \int_{\frac{2}{5}}^{\frac{4}{5}} \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}{5} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{2}{5}}^{\frac{4}{5}})

Vereinfache

\frac{1}{5} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{2}{5}}^{\frac{4}{5}}) : - \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{2}{5}}^{\frac{4}{5}}

= - \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{2}{5}}^{\frac{4}{5}}

Berechne die Grenzen:

\frac{ln ( \frac{9}{5} ) + ln ( 5 ) - ln ( \frac{7}{5} ) + ln ( \frac{3}{5} )}{2}

= - \frac{1}{5} \cdot \frac{ln ( \frac{9}{5} ) + ln ( 5 ) - ln ( \frac{7}{5} ) + ln ( \frac{3}{5} )}{2}

Vereinfache

= - \frac{2 ln ( 3 ) - ln ( \frac{7}{5} ) + ln ( \frac{3}{5} )}{10}

x \to 0 lim ((4 x)^{7 x})

a re a - 2 x^{2} + x, - 2, 1

n = 0 \sum \infty \frac{1}{2}

\frac{\partial}{\partial x} (a sin (2 x))

\int \frac{e ^{7 x}}{7 x} d x