What is the integral from 1 to 6 of x^3ln(4x) ?

The integral from 1 to 6 of x^3ln(4x) is 324ln(24)-1295/16-1/2 ln(2)

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integral from 1 to 6 of x^3ln(4x)

\int_{1}^{6} x^{3} ln (4 x) d x

324 ln (24) - \frac{1295}{16} - \frac{1}{2} ln (2)

Decimal Notation

948.40536 \dots

Solution steps

\int_{1}^{6} 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]_{1}^{6}

= [\frac{1}{4} x^{4} ln (4 x) - \int \frac{x ^{3}}{4} d x]_{1}^{6}

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

= [\frac{1}{4} x^{4} ln (4 x) - \frac{x ^{4}}{16}]_{1}^{6}

Compute the boundaries:

324 ln (24) - \frac{1295}{16} - \frac{1}{2} ln (2)

= 324 ln (24) - \frac{1295}{16} - \frac{1}{2} ln (2)

\frac{( s + 1 )}{( s + 2 ) ^{2}}

\int \frac{( ln ( x ) ) ^{22}}{x} d x

\int x (15 x + 6) d x

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

\frac{d}{d t} (cos^{3} (t) + sin^{3} (t))

What is the integral from 1 to 6 of x^3ln(4x) ?
The integral from 1 to 6 of x^3ln(4x) is 324ln(24)-1295/16-1/2 ln(2)