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

The integral from 2 to 8 of t^3ln(2t) is 4088ln(2)-255

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

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

4088 ln (2) - 255

Decimal Notation

2578.58567 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

= [\frac{1}{4} t^{4} ln (2 t) - \frac{t ^{4}}{16}]_{2}^{8}

Compute the boundaries:

4088 ln (2) - 255

= 4088 ln (2) - 255

\frac{\partial}{\partial x} (3 x + 4 y)

15 - x^{2}, x^{2} - 3, - 3, 3

4 + 0.28 t^{- 1}

x \to 1 lim (5 x - 2)

\frac{d}{d x} ([x + (x + sin^{2} (x))^{6}]^{5})

What is the integral from 2 to 8 of t^3ln(2t) ?
The integral from 2 to 8 of t^3ln(2t) is 4088ln(2)-255