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

Lösung

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

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

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

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

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

Vereinfache

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

\int_{0}^{3} 2 t d t

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