integral of sqrt(x^2+10)

Lösung

\int x^{2} + 10 d x

10 (\frac{1}{20} x 10 + x^{2} + \frac{1}{2} ln \frac{1}{10} x + \frac{1}{10} (10 + x^{2})) + C

Schritte zur Lösung

\int x^{2} + 10 d x

Trigonometrische Substitution anwenden

= \int 10 sec^{3} (u) d u

Entferne die Konstante:

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

= 10 \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= 10 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 10 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arctan (\frac{1}{10} x)

ein

= 10 \frac{sec ^{2} ( arctan ( \frac{1}{10} x ) ) sin ( arctan ( \frac{1}{10} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x))

Vereinfache

10 \frac{sec ^{2} ( arctan ( \frac{1}{10} x ) ) sin ( arctan ( \frac{1}{10} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x)) : 10 (\frac{1}{20} x 10 + x^{2} + \frac{1}{2} ln \frac{1}{10} x + \frac{1}{10} (10 + x^{2}))

= 10 (\frac{1}{20} x 10 + x^{2} + \frac{1}{2} ln \frac{1}{10} x + \frac{1}{10} (10 + x^{2}))

Füge eine Konstante zur Lösung hinzu

= 10 (\frac{1}{20} x 10 + x^{2} + \frac{1}{2} ln \frac{1}{10} x + \frac{1}{10} (10 + x^{2})) + C

(sin (x))^{^{'}}

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