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

Lösung

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

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

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

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

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

Vereinfache

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

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