What is the integral from 3 to 6 of t^4ln(2t) ?

The integral from 3 to 6 of t^4ln(2t) is -7533/25+(-243ln(6)+7776ln(12))/5

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integral from 3 to 6 of t^4ln(2t)

\int_{3}^{6} t^{4} ln (2 t) d t

- \frac{7533}{25} + \frac{- 243 ln ( 6 ) + 7776 ln ( 12 )}{5}

Decimal Notation

3476.12731 \dots

Solution steps

\int_{3}^{6} t^{4} ln (2 t) d t

Apply Integration By Parts

: [\frac{1}{5} t^{5} ln (2 t) - \int \frac{t ^{4}}{5} d t]_{3}^{6}

= [\frac{1}{5} t^{5} ln (2 t) - \int \frac{t ^{4}}{5} d t]_{3}^{6}

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

= [\frac{1}{5} t^{5} ln (2 t) - \frac{t ^{5}}{25}]_{3}^{6}

Compute the boundaries:

- \frac{7533}{25} + \frac{- 243 ln ( 6 ) + 7776 ln ( 12 )}{5}

= - \frac{7533}{25} + \frac{- 243 ln ( 6 ) + 7776 ln ( 12 )}{5}

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What is the integral from 3 to 6 of t^4ln(2t) ?
The integral from 3 to 6 of t^4ln(2t) is -7533/25+(-243ln(6)+7776ln(12))/5