integral of ((y^3+4y))/(9+y^4)

Lösung

\int \frac{( y ^{3} + 4 y )}{9 + y ^{4}} d y

\frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3}) + C

Schritte zur Lösung

\int \frac{y ^{3} + 4 y}{9 + y ^{4}} d y

Multipliziere aus

\frac{y ^{3} + 4 y}{9 + y ^{4}} : \frac{y ^{3}}{9 + y ^{4}} + \frac{4 y}{9 + y ^{4}}

Wende die Summenregel an:

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

= \int \frac{y ^{3}}{9 + y ^{4}} d y + \int \frac{4 y}{9 + y ^{4}} d y

\int \frac{y ^{3}}{9 + y ^{4}} d y = \frac{1}{4} ln 9 + y^{4}

\int \frac{4 y}{9 + y ^{4}} d y = \frac{2}{3} arctan (\frac{y ^{2}}{3})

= \frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3}) + C

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