integral of x^2sqrt(x^2+4)

Lösung

\int x^{2} x^{2} + 4 d x

16 (- \frac{1}{32} (x 4 + x^{2} + 4 ln (\frac{x + 4 + x ^{2}}{2})) + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}}) + C

Schritte zur Lösung

\int x^{2} x^{2} + 4 d x

Trigonometrische Substitution anwenden

= \int 16 sec^{3} (u) tan^{2} (u) d u

Entferne die Konstante:

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

= 16 \cdot \int sec^{3} (u) tan^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 16 \cdot \int sec^{3} (u) (- 1 + sec^{2} (u)) d u

Multipliziere aus

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= 16 \cdot \int - sec^{3} (u) + sec^{5} (u) d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 16 (- \int sec^{3} (u) d u + \int sec^{5} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= 16 (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Setze in

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

ein

= 16 (- (\frac{1}{2} sec (arctan (\frac{1}{2} x)) tan (arctan (\frac{1}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x))) + \frac{sec ^{4} ( arctan ( \frac{1}{2} x ) ) sin ( arctan ( \frac{1}{2} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{2} x)) tan (arctan (\frac{1}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x))))

Vereinfache

16 (- (\frac{1}{2} sec (arctan (\frac{1}{2} x)) tan (arctan (\frac{1}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x))) + \frac{sec ^{4} ( arctan ( \frac{1}{2} x ) ) sin ( arctan ( \frac{1}{2} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{2} x)) tan (arctan (\frac{1}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x)))) : 16 (- \frac{1}{32} (x 4 + x^{2} + 4 ln (\frac{x + 4 + x ^{2}}{2})) + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}})

= 16 (- \frac{1}{32} (x 4 + x^{2} + 4 ln (\frac{x + 4 + x ^{2}}{2})) + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}})

Füge eine Konstante zur Lösung hinzu

= 16 (- \frac{1}{32} (x 4 + x^{2} + 4 ln (\frac{x + 4 + x ^{2}}{2})) + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}}) + C

\int \frac{x ^{4}}{1 - x ^{10}} d x

x \to \infty lim (e x)

\frac{\partial}{\partial y} (x^{2} \cdot y^{2})

y^{^{''}} - 7 y^{^{'}} - 18 y = 0

\frac{\partial}{\partial y} (y^{2} cos (x))