What is the integral of y^3ln^3(y) ?

The integral of y^3ln^3(y) is 1/4 y^4ln^3(y)-3/4 (1/4 y^4ln^2(y)-1/2 (1/4 y^4ln(y)-(y^4)/(16)))+C

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integral of y^3ln^3(y)

\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

Solution steps

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

Apply Integration By Parts

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

= \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}))

Add a constant to the solution

= \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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What is the integral of y^3ln^3(y) ?
The integral of y^3ln^3(y) is 1/4 y^4ln^3(y)-3/4 (1/4 y^4ln^2(y)-1/2 (1/4 y^4ln(y)-(y^4)/(16)))+C