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

The integral from 1 to 4 of t^4ln(3t) is 1024/5 ln(12)-1024/25-(1/5 ln(3)-1/25)

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integral from 1 to 4 of t^4ln(3t)

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

\frac{1024}{5} ln (12) - \frac{1024}{25} - (\frac{1}{5} ln (3) - \frac{1}{25})

Decimal Notation

467.76915 \dots

Solution steps

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

Compute the indefinite integral:

\int t^{4} ln (3 t) d t = \frac{1}{5} t^{5} ln (3 t) - \frac{t ^{5}}{25} + C

Compute the boundaries:

\int_{1}^{4} t^{4} ln (3 t) d t = \frac{1024}{5} ln (12) - \frac{1024}{25} - (\frac{1}{5} ln (3) - \frac{1}{25})

= \frac{1024}{5} ln (12) - \frac{1024}{25} - (\frac{1}{5} ln (3) - \frac{1}{25})

\int (x^{2} - 2 x + 1) e^{2 x} d x

\frac{\partial}{\partial x} (2 x y - cos (x + y) - e^{y})

n = 5 \sum \infty ln (n) - ln (2 n)

\frac{d}{d x} (- (cos (x)) ln (sec (x) + tan (x)))

x \to π lim (2 cot (x))

What is the integral from 1 to 4 of t^4ln(3t) ?
The integral from 1 to 4 of t^4ln(3t) is 1024/5 ln(12)-1024/25-(1/5 ln(3)-1/25)