integral of y^3ln^3(y)

Lösung

\int y^{3} ln^{3} (y) d y

\frac{1}{4} y^{4} ln^{3} (y) - \frac{3}{4} (\frac{1}{4} y^{4} ln^{2} (y) - \frac{1}{2} (\frac{1}{4} y^{4} ln (y) - \frac{y ^{4}}{16})) + C

Schritte zur Lösung

\int y^{3} ln^{3} (y) d y

Wende die partielle Integration an

= \frac{1}{4} y^{4} ln^{3} (y) - \int \frac{3}{4} y^{3} ln^{2} (y) d y

\int \frac{3}{4} y^{3} ln^{2} (y) d y = \frac{3}{4} (\frac{1}{4} y^{4} ln^{2} (y) - \frac{1}{2} (\frac{1}{4} y^{4} ln (y) - \frac{y ^{4}}{16}))

= \frac{1}{4} y^{4} ln^{3} (y) - \frac{3}{4} (\frac{1}{4} y^{4} ln^{2} (y) - \frac{1}{2} (\frac{1}{4} y^{4} ln (y) - \frac{y ^{4}}{16}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} y^{4} ln^{3} (y) - \frac{3}{4} (\frac{1}{4} y^{4} ln^{2} (y) - \frac{1}{2} (\frac{1}{4} y^{4} ln (y) - \frac{y ^{4}}{16})) + C

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