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

Lösung

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

- 5 + \frac{x ^{2} 1 + x ^{4} + ln x ^{2} + 1 + x ^{4} - ln ( 2 + 5 )}{2}

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

\int_{a r c t a n (2)}^{a r c t a n (x^{2})} sec (u) d u = ln x^{2} + 1 + x^{4} - ln (2 + 5)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - 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^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (2 + 5)) : [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (2 + 5))

= [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (x^{2})} + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (2 + 5))

Berechne die Grenzen:

\frac{1}{2} x^{2} 1 + x^{4} - 5

= \frac{1}{2} x^{2} 1 + x^{4} - 5 + \frac{1}{2} (ln x^{2} + 1 + x^{4} - ln (2 + 5))

Vereinfache

= - 5 + \frac{x ^{2} 1 + x ^{4} + ln x ^{2} + 1 + x ^{4} - ln ( 2 + 5 )}{2}

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