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

Lösung

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

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

Dezimale

1.22201 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Entferne die Konstante:

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

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

Wende die partielle Integration an

= - - [- \frac{1 + u ^{2}}{u} - \int - \frac{1}{1 + u ^{2}} d u]_{\frac{1}{2}}^{1}

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

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

Vereinfache

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

Berechne die Grenzen:

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

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

Vereinfache

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

\int_{0}^{\frac{3 π}{2}} 5 sin (x) d x

\int_{0}^{2} (x + 1) (3 x - 4) d x

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

\int_{0}^{π} 7 e^{x} + 5 sin (x) d x

\int_{1}^{e} \frac{4 ln ( x )}{x} d x