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

Lösung

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

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

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Multipliziere aus

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

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

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

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

= a^{2} (- ln ∣ tan (u) + sec (u) ∣ + \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^{2} (- ln tan (arctan (\frac{1}{a} x)) + sec (arctan (\frac{1}{a} x)) + \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^{2} (- ln tan (arctan (\frac{1}{a} x)) + sec (arctan (\frac{1}{a} x)) + \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{1}{2} x x^{2} + a^{2} - \frac{a ^{2}}{2} ln (\frac{x + x ^{2} + a ^{2}}{∣ a ∣})

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

Füge eine Konstante zur Lösung hinzu

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

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in t e g r a l csc (x)

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