integral (r^2)/(sqrt(1+r^2))

Lösung

integral

\frac{r ^{2}}{1 + r ^{2}}

\frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1 + C

Schritte zur Lösung

\int \frac{r ^{2}}{1 + r ^{2}} d r

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Multipliziere aus

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

= \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

= - \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) ∣

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

Setze in

u = arctan (r)

ein

= - ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ + \frac{1}{2} sec (arctan (r)) tan (arctan (r)) + \frac{1}{2} ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣

Vereinfache

- ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ + \frac{1}{2} sec (arctan (r)) tan (arctan (r)) + \frac{1}{2} ln ∣ tan (arctan (r)) + sec (arctan (r)) ∣ : \frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1

= \frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} r r^{2} + 1 - \frac{1}{2} ln r + r^{2} + 1 + C

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

in t e g r a l \frac{x}{x ^{2}}

in t e g r a l \frac{1}{x ^{2} + 4 x + 9}

in t e g r a l \frac{1}{x ^{6}}

in t e g r a l 2 e^{u}