integral from 9/8 to 4 of 1/(1-x^2)

Lösung

\int_{\frac{9}{8}}^{4} \frac{1}{1 - x ^{2}} d x

\frac{ln ( 5 ) - ln ( 51 )}{2}

Dezimale

- 1.16119 \dots

Schritte zur Lösung

\int_{\frac{9}{8}}^{4} \frac{1}{1 - x ^{2}} d x

Nutze das gemeinsame Integral :

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

= [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{9}{8}}^{4}

Vereinfache

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{9}{8}}^{4} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{9}{8}}^{4}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{9}{8}}^{4}

Berechne die Grenzen:

\frac{ln ( 5 ) - ln ( 3 ) - ln ( \frac{17}{8} ) - 3 ln ( 2 )}{2}

= \frac{ln ( 5 ) - ln ( 3 ) - ln ( \frac{17}{8} ) - 3 ln ( 2 )}{2}

Vereinfache

= \frac{ln ( 5 ) - ln ( 51 )}{2}

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

d er i v a t i v e e^{- 1}

\int_{0}^{2} e^{- x} + e d x

\int t e^{- 6 t} d t

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 6}{x})