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

Lösung

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

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

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{a r c t a n (x)}^{\frac{π}{4}} sec^{3} (u) d u

Wende integrale Reduktion an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

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

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

Vereinfache

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

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