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

The integral from 3 to 8 of t^3ln(2t) is 4096ln(2)-4015/16-81/4 ln(6)

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

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

4096 ln (2) - \frac{4015}{16} - \frac{81}{4} ln (6)

Decimal Notation

2551.91022 \dots

Solution steps

\int_{3}^{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]_{3}^{8}

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

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

= [\frac{1}{4} t^{4} ln (2 t) - \frac{t ^{4}}{16}]_{3}^{8}

Compute the boundaries:

4096 ln (2) - \frac{4015}{16} - \frac{81}{4} ln (6)

= 4096 ln (2) - \frac{4015}{16} - \frac{81}{4} ln (6)

\frac{1}{9 s - 1}

x \to 0 lim (\frac{cos ( 2 x )}{x})

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

\frac{\partial}{\partial x} (e^{(\frac{x}{y})})

\int 2 x^{4} x^{5} + 6 d x

What is the integral from 3 to 8 of t^3ln(2t) ?
The integral from 3 to 8 of t^3ln(2t) is 4096ln(2)-4015/16-81/4 ln(6)