integral from 0 to x of sqrt(t^2+4)

Lösung

\int_{0}^{x} t^{2} + 4 d t

4 (\frac{x x ^{2} + 4}{8} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

Schritte zur Lösung

\int_{0}^{x} t^{2} + 4 d t

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (\frac{x}{2})} 4 sec^{3} (u) d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{a r c t a n (\frac{x}{2})} sec^{3} (u) d u

Wende integrale Reduktion an

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{x}{2})} sec (u) d u

\int_{0}^{a r c t a n (\frac{x}{2})} sec (u) d u = ln (\frac{x + x ^{2} + 4}{2})

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2})

Vereinfache

4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}) : 4 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

= 4 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

Berechne die Grenzen:

\frac{x x ^{2} + 4}{8}

= 4 (\frac{x x ^{2} + 4}{8} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

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