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

Lösung

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

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

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

\int_{a r c t a n (\frac{x}{2})}^{0} sec (u) d u = - ln (\frac{1}{2} x + x^{2} + 4)

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

Vereinfache

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

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

Berechne die Grenzen:

- \frac{1}{8} x x^{2} + 4

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

Vereinfache

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

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