integral sqrt(x^2+2)

Lösung

integral

x^{2} + 2

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

Schritte zur Lösung

\int x^{2} + 2 d x

Trigonometrische Substitution anwenden

= \int 2 sec^{3} (u) d u

Entferne die Konstante:

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

= 2 \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= 2 (\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) ∣

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

Setze in

u = arctan (\frac{1}{2} x)

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{2 x}{( 2 x + 1 )}

in t e g r a l \frac{1}{x ^{2}} lo g_{10} (x)

in t e g r a l 2 t + 3

in t e g r a l \frac{( x ^{2} + 1 ) e ^{x}}{( x + 1 ) ^{2}}

in t e g r a l e^{6 x}