integral from x to 0 of sqrt(t^2+16)

Lösung

\int_{x}^{0} t^{2} + 16 d t

16 (- \frac{x x ^{2} + 16}{32} - \frac{1}{2} ln (\frac{x + x ^{2} + 16}{4}))

Schritte zur Lösung

\int_{x}^{0} t^{2} + 16 d t

Trigonometrische Substitution anwenden

= \int_{a r c t a n (\frac{x}{4})}^{0} 16 sec^{3} (u) d u

Entferne die Konstante:

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

= 16 \cdot \int_{a r c t a n (\frac{x}{4})}^{0} sec^{3} (u) d u

Wende integrale Reduktion an

= 16 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{x}{4})}^{0} + \frac{1}{2} \cdot \int_{a r c t a n (\frac{x}{4})}^{0} sec (u) d u)

\int_{a r c t a n (\frac{x}{4})}^{0} sec (u) d u = - ln (\frac{x + x ^{2} + 16}{4})

= 16 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{x}{4})}^{0} + \frac{1}{2} (- ln (\frac{x + x ^{2} + 16}{4})))

Vereinfache

16 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (\frac{x}{4})}^{0} + \frac{1}{2} (- ln (\frac{x + x ^{2} + 16}{4}))) : 16 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{x}{4})}^{0} - \frac{1}{2} ln (\frac{x + x ^{2} + 16}{4}))

= 16 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{x}{4})}^{0} - \frac{1}{2} ln (\frac{x + x ^{2} + 16}{4}))

Berechne die Grenzen:

- \frac{x x ^{2} + 16}{32}

= 16 (- \frac{x x ^{2} + 16}{32} - \frac{1}{2} ln (\frac{x + x ^{2} + 16}{4}))

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