What is the integral from 4 to 8 of t^2ln(3t) ?

The integral from 4 to 8 of t^2ln(3t) is -448/9+(512ln(24)-64ln(12))/3

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integral from 4 to 8 of t^2ln(3t)

\int_{4}^{8} t^{2} ln (3 t) d t

- \frac{448}{9} + \frac{512 ln ( 24 ) - 64 ln ( 12 )}{3}

Decimal

439.59873 \dots

Solution steps

\int_{4}^{8} t^{2} ln (3 t) d t

Apply Integration By Parts

= [\frac{1}{3} t^{3} ln (3 t) - \int \frac{t ^{2}}{3} d t]_{4}^{8}

\int \frac{t ^{2}}{3} d t = \frac{t ^{3}}{9}

= [\frac{1}{3} t^{3} ln (3 t) - \frac{t ^{3}}{9}]_{4}^{8}

Compute the boundaries:

- \frac{448}{9} + \frac{512 ln ( 24 ) - 64 ln ( 12 )}{3}

= - \frac{448}{9} + \frac{512 ln ( 24 ) - 64 ln ( 12 )}{3}

\int \frac{1}{6 x ( 3 + x )} d x

(x^{2} - x y + 2 y^{2}) d x + (x^{2} - 2 x y) d y = 0

(\frac{1}{1 + x})^{^{'}}

x \to 0 lim (1 + csc (x))

\int (sec (x) + tan (x)) d x

What is the integral from 4 to 8 of t^2ln(3t) ?
The integral from 4 to 8 of t^2ln(3t) is -448/9+(512ln(24)-64ln(12))/3