integral from 3 to x of sqrt(t^2+1)

Lösung

\int_{3}^{x} t^{2} + 1 d t

\frac{x 1 + x ^{2}}{2} - \frac{3 5}{2} + \frac{1}{2} (ln x + 1 + x^{2} - ln (3 + 10))

Schritte zur Lösung

\int_{3}^{x} t^{2} + 1 d t

Trigonometrische Substitution anwenden

= \int_{a r c t a n (3)}^{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 (3)}^{a r c t a n (x)} + \frac{1}{2} \cdot \int_{a r c t a n (3)}^{a r c t a n (x)} sec (u) d u

\int_{a r c t a n (3)}^{a r c t a n (x)} sec (u) d u = ln x + 1 + x^{2} - ln (3 + 10)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (3)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (3 + 10))

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (3)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (3 + 10)) : [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (3)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (3 + 10))

= [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (3)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (3 + 10))

Berechne die Grenzen:

\frac{1}{2} x 1 + x^{2} - \frac{3 5}{2}

= \frac{1}{2} x 1 + x^{2} - \frac{3 5}{2} + \frac{1}{2} (ln x + 1 + x^{2} - ln (3 + 10))

Vereinfache

= \frac{x 1 + x ^{2}}{2} - \frac{3 5}{2} + \frac{1}{2} (ln x + 1 + x^{2} - ln (3 + 10))

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