What is the integral from 1 to 2 of x^4ln(3x) ?

The integral from 1 to 2 of x^4ln(3x) is -31/25+(ln(2^{32}*3^{31}))/5

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integral from 1 to 2 of x^4ln(3x)

\int_{1}^{2} x^{4} ln (3 x) d x

- \frac{31}{25} + \frac{ln ( 2 ^{32} \cdot 3 ^{31} )}{5}

Decimal Notation

10.00753 \dots

Solution steps

\int_{1}^{2} x^{4} ln (3 x) d x

Apply Integration By Parts

: [\frac{1}{5} x^{5} ln (3 x) - \int \frac{x ^{4}}{5} d x]_{1}^{2}

= [\frac{1}{5} x^{5} ln (3 x) - \int \frac{x ^{4}}{5} d x]_{1}^{2}

\int \frac{x ^{4}}{5} d x = \frac{x ^{5}}{25}

= [\frac{1}{5} x^{5} ln (3 x) - \frac{x ^{5}}{25}]_{1}^{2}

Compute the boundaries:

- \frac{31}{25} + \frac{32 ln ( 6 ) - ln ( 3 )}{5}

= - \frac{31}{25} + \frac{32 ln ( 6 ) - ln ( 3 )}{5}

Simplify

= - \frac{31}{25} + \frac{ln ( 2 ^{32} \cdot 3 ^{31} )}{5}

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What is the integral from 1 to 2 of x^4ln(3x) ?
The integral from 1 to 2 of x^4ln(3x) is -31/25+(ln(2^{32}*3^{31}))/5