integral of 1/(y^3sqrt(y^2+1))

Lösung

\int \frac{1}{y ^{3} y ^{2} + 1} d y

\frac{1}{8} (tan^{2} (\frac{arctan ( y )}{2}) - 4 ln tan (\frac{arctan ( y )}{2}) - cot^{2} (\frac{arctan ( y )}{2})) + C

Schritte zur Lösung

\int \frac{1}{y ^{3} y ^{2} + 1} d y

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int \frac{cos ^{2} ( u )}{sin ^{3} ( u )} d u

Wende U-Substitution an

= \int \frac{( - v ^{2} + 1 ) ^{2}}{4 v ^{3}} d v

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int \frac{( - v ^{2} + 1 ) ^{2}}{v ^{3}} d v

Multipliziere aus

\frac{( - v ^{2} + 1 ) ^{2}}{v ^{3}} : v - \frac{2}{v} + \frac{1}{v ^{3}}

= \frac{1}{4} \cdot \int v - \frac{2}{v} + \frac{1}{v ^{3}} d v

Wende die Summenregel an:

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

= \frac{1}{4} (\int v d v - \int \frac{2}{v} d v + \int \frac{1}{v ^{3}} d v)

\int v d v = \frac{v ^{2}}{2}

\int \frac{2}{v} d v = 2 ln ∣ v ∣

\int \frac{1}{v ^{3}} d v = - \frac{1}{2 v ^{2}}

= \frac{1}{4} (\frac{v ^{2}}{2} - 2 ln ∣ v ∣ - \frac{1}{2 v ^{2}})

Ersetze zurück

= \frac{1}{4} \frac{tan ^{2} ( \frac{a r c t a n ( y )}{2} )}{2} - 2 ln tan (\frac{arctan ( y )}{2}) - \frac{1}{2 tan ^{2} ( \frac{a r c t a n ( y )}{2} )}

Vereinfache

\frac{1}{4} \frac{tan ^{2} ( \frac{a r c t a n ( y )}{2} )}{2} - 2 ln tan (\frac{arctan ( y )}{2}) - \frac{1}{2 tan ^{2} ( \frac{a r c t a n ( y )}{2} )} : \frac{1}{8} (tan^{2} (\frac{arctan ( y )}{2}) - 4 ln tan (\frac{arctan ( y )}{2}) - cot^{2} (\frac{arctan ( y )}{2}))

= \frac{1}{8} (tan^{2} (\frac{arctan ( y )}{2}) - 4 ln tan (\frac{arctan ( y )}{2}) - cot^{2} (\frac{arctan ( y )}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} (tan^{2} (\frac{arctan ( y )}{2}) - 4 ln tan (\frac{arctan ( y )}{2}) - cot^{2} (\frac{arctan ( y )}{2})) + C

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