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

The integral from 2 to 6 of t^4ln(3t) is -7744/25+(-32ln(6)+7776ln(18))/5

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

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

- \frac{7744}{25} + \frac{- 32 ln ( 6 ) + 7776 ln ( 18 )}{5}

Decimal Notation

4173.87889 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

- \frac{7744}{25} + \frac{- 32 ln ( 6 ) + 7776 ln ( 18 )}{5}

= - \frac{7744}{25} + \frac{- 32 ln ( 6 ) + 7776 ln ( 18 )}{5}

f (x) = x^{2} - 16

y = \frac{3}{x} + 2 x^{3}, 2 \leq x \leq 3

f (x) = - 2 x^{3}, (1, - 2)

\frac{d}{d x} (x (x + 12))

\frac{\partial}{\partial x} (3 z cos (4 x y) + x - 7 y)

What is the integral from 2 to 6 of t^4ln(3t) ?
The integral from 2 to 6 of t^4ln(3t) is -7744/25+(-32ln(6)+7776ln(18))/5