English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

Lösung

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

- \frac{a ^{2} x x ^{2} + a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + \frac{1}{4} x (x^{2} + a^{2})^{\frac{3}{2}} + C

Schritte zur Lösung

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

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

= \int x^{2} a^{2} (1 + \frac{x ^{2}}{a ^{2}}) d x

Trigonometrische Substitution anwenden

= \int a^{4} 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

= a^{4} \cdot \int sec^{3} (u) tan^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= a^{4} \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)

= a^{4} \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

= a^{4} (- \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) ∣)

= a^{4} (- (\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}{a} x)

ein

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

Vereinfache

a^{4} (- (\frac{1}{2} sec (arctan (\frac{1}{a} x)) tan (arctan (\frac{1}{a} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{a} x)) + sec (arctan (\frac{1}{a} x))) + \frac{sec ^{4} ( arctan ( \frac{1}{a} x ) ) sin ( arctan ( \frac{1}{a} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{a} x)) tan (arctan (\frac{1}{a} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{a} x)) + sec (arctan (\frac{1}{a} x)))) : - \frac{a ^{2} x x ^{2} + a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + \frac{1}{4} x (x^{2} + a^{2})^{\frac{3}{2}}

= - \frac{a ^{2} x x ^{2} + a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + \frac{1}{4} x (x^{2} + a^{2})^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= - \frac{a ^{2} x x ^{2} + a ^{2}}{8} - \frac{a ^{4}}{8} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + \frac{1}{4} x (x^{2} + a^{2})^{\frac{3}{2}} + C

\int \frac{1}{1 - 16 y ^{2}} d y

\int e^{2 y} (4 - e^{2 y})^{\frac{4}{3}} d y

\int (e^{y^{2}} y) d y

\int \frac{- 3 x ^{2}}{16 - x ^{2}} d x

\int 80 d x