What is the integral from 1 to 5 of t^3ln(4t) ?

The integral from 1 to 5 of t^3ln(4t) is -39-1/2 ln(2)+625/4 ln(20)

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integral from 1 to 5 of t^3ln(4t)

\int_{1}^{5} t^{3} ln (4 t) d t

- 39 - \frac{1}{2} ln (2) + \frac{625}{4} ln (20)

Decimal Notation

428.73659 \dots

Solution steps

\int_{1}^{5} t^{3} ln (4 t) d t

Apply Integration By Parts

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

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

\int \frac{t ^{3}}{4} d t = \frac{t ^{4}}{16}

= [\frac{1}{4} t^{4} ln (4 t) - \frac{t ^{4}}{16}]_{1}^{5}

Compute the boundaries:

- 39 - \frac{1}{2} ln (2) + \frac{625}{4} ln (20)

= - 39 - \frac{1}{2} ln (2) + \frac{625}{4} ln (20)

\int \frac{x - 1}{x ^{2} - 2 x + 3} d x

3 x^{- \frac{1}{2}}

\frac{d}{d x} ((7 arcsin (4 x))^{x})

\frac{\partial}{\partial x} (\frac{z x - y}{x})

\frac{d}{d x} (3^{x} - 1)

What is the integral from 1 to 5 of t^3ln(4t) ?
The integral from 1 to 5 of t^3ln(4t) is -39-1/2 ln(2)+625/4 ln(20)