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Beliebt Rechnen >

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

Lösung

\int_{2}^{x} 1 + t^{2} d t

Lösung

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

Schritte zur Lösung

\int_{2}^{x} 1 + t^{2} d t

Trigonometrische Substitution anwenden

= \int_{a r c t a n (2)}^{a r c t a n (x)} sec^{3} (u) d u

Wende integrale Reduktion an

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} \cdot \int_{a r c t a n (2)}^{a r c t a n (x)} sec (u) d u

\int_{a r c t a n (2)}^{a r c t a n (x)} sec (u) d u = ln x + 1 + x^{2} - ln (2 + 5)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5))

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5)) : [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5))

= [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5))

Berechne die Grenzen:

\frac{1}{2} x 1 + x^{2} - 5

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

Vereinfache

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

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