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

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

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integral from 1 to e of x^3ln(3x)

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

\frac{4 e ^{4} ln ( 3 e ) - e ^{4} - 4 ln ( 3 ) + 1}{16}

Decimal

25.02054 \dots

Solution steps

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

Apply Integration By Parts

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

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

= [\frac{1}{4} (x^{4} ln (3 x) - 4 \cdot \frac{x ^{4}}{16})]_{1}^{e}

4 \cdot \frac{x ^{4}}{16} = \frac{x ^{4}}{4}

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

Compute the boundaries:

\frac{4 e ^{4} ln ( 3 e ) - e ^{4} - 4 ln ( 3 ) + 1}{16}

= \frac{4 e ^{4} ln ( 3 e ) - e ^{4} - 4 ln ( 3 ) + 1}{16}

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