What is the integral of x^3ln(4x) ?

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

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integral of x^3ln(4x)

\int x^{3} ln (4 x) d x

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

Solution steps

\int x^{3} ln (4 x) d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\int sinh (x) d x

\frac{d}{d x} (arcsin (\frac{1}{x ^{2} + 4}))

\int 1 + x^{\frac{4}{5}} d x

y = \frac{1 - x ^{2}}{( 1 + x ^{2} ) ^{2}}

f (x) = 1 - 2 x

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