integral from 2 to 5 of 1/(xsqrt(x^2+1))

Lösung

\int_{2}^{5} \frac{1}{x x ^{2} + 1} d x

- \frac{ln ( 26 + 1 ) - ln ( ( 26 - 1 ) ( 5 + 1 ) ) + ln ( 5 - 1 )}{2}

Dezimale

0.28252 \dots

Schritte zur Lösung

\int_{2}^{5} \frac{1}{x x ^{2} + 1} d x

Wende U-Substitution an

= \int_{5}^{26} \frac{1}{u ^{2} - 1} d u

Faktorisiere

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

= \int_{5}^{26} \frac{1}{- ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= - \int_{5}^{26} \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{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{5}^{26}

Vereinfache

- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{5}^{26} : - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{5}^{26}

= - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{5}^{26}

Berechne die Grenzen:

\frac{ln ( 26 + 1 ) - ln ( 26 - 1 ) - ln ( 5 + 1 ) + ln ( 5 - 1 )}{2}

= - \frac{ln ( 26 + 1 ) - ln ( 26 - 1 ) - ln ( 5 + 1 ) + ln ( 5 - 1 )}{2}

Vereinfache

= - \frac{ln ( 26 + 1 ) - ln ( ( 26 - 1 ) ( 5 + 1 ) ) + ln ( 5 - 1 )}{2}

\int_{- 1}^{1} \frac{x + 3}{x ^{2} + 2 x + 5} d x

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

\int_{0}^{t} e^{- u} (t - u)^{2} d u

\int_{0}^{\frac{π}{4}} (7 sec^{2} (x)) d x

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