integral of x^2sqrt(x^2+1)

Lösung

\int x^{2} x^{2} + 1 d x

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

Schritte zur Lösung

\int x^{2} x^{2} + 1 d x

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \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)

= \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

= - \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) ∣)

= - (\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 (x)

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

n or ma l - x^{3} + 2 x^{2} - 2, at x = 1

\int - x^{5} e^{x} d x

\frac{\partial}{\partial x} (4 y x^{3} - 6 y^{2} x^{2})

t an g e n t f (x) = \frac{1}{1 + x ^{2}}, (1, \frac{1}{2})

\frac{d}{d y} (4)