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

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

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integral of z^3ln(z)

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

\frac{1}{4} z^{4} ln (z) - \frac{z ^{4}}{16} + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{z ^{3}}{4} d z = \frac{z ^{4}}{16}

= \frac{1}{4} z^{4} ln (z) - \frac{z ^{4}}{16}

Add a constant to the solution

= \frac{1}{4} z^{4} ln (z) - \frac{z ^{4}}{16} + C

\int (2 x + 4) \cdot e^{2 x + 4} d x

\int \frac{( 3 x - 2 )}{x + 1} d x

\frac{d}{d x} (x + x^{2} + x^{3})

\frac{\partial}{\partial x} (sin (y) - cos (x))

y = \frac{1}{x}, y = \frac{1}{x ^{2}}, x = 5

What is the integral of z^3ln(z) ?
The integral of z^3ln(z) is 1/4 z^4ln(z)-(z^4)/(16)+C