What is the integral from 1 to 6 of t^4ln(2t) ?

The integral from 1 to 6 of t^4ln(2t) is -311-1/5 ln(2)+7776/5 ln(12)

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integral from 1 to 6 of t^4ln(2t)

\int_{1}^{6} t^{4} ln (2 t) d t

- 311 - \frac{1}{5} ln (2) + \frac{7776}{5} ln (12)

Decimal Notation

3553.38819 \dots

Solution steps

\int_{1}^{6} t^{4} ln (2 t) d t

Compute the indefinite integral:

\int t^{4} ln (2 t) d t = \frac{1}{5} t^{5} ln (2 t) - \frac{t ^{5}}{25} + C

Compute the boundaries:

\int_{1}^{6} t^{4} ln (2 t) d t = \frac{7776}{5} ln (12) - \frac{7776}{25} - (\frac{1}{5} ln (2) - \frac{1}{25})

= \frac{7776}{5} ln (12) - \frac{7776}{25} - (\frac{1}{5} ln (2) - \frac{1}{25})

Simplify

= - 311 - \frac{1}{5} ln (2) + \frac{7776}{5} ln (12)

\frac{d}{d x} (ln (ln (\frac{1}{x})))

y^{^{'}} + y = 2

\int \frac{1}{15 + 4 t - t ^{2}} d t

\frac{1}{1 - x}, 22

\frac{s}{s ^{2} + 2}

What is the integral from 1 to 6 of t^4ln(2t) ?
The integral from 1 to 6 of t^4ln(2t) is -311-1/5 ln(2)+7776/5 ln(12)