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

The integral from 1 to 4 of t^2ln(4t) is -7+254/3 ln(2)

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

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

- 7 + \frac{254}{3} ln (2)

Decimal

51.68646 \dots

Solution steps

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

Apply Integration By Parts

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

\int \frac{t ^{2}}{3} d t = \frac{t ^{3}}{9}

= [\frac{1}{3} t^{3} ln (4 t) - \frac{t ^{3}}{9}]_{1}^{4}

Compute the boundaries:

- 7 + \frac{254}{3} ln (2)

= - 7 + \frac{254}{3} ln (2)

y^{^{'}} + 6 y = t + e^{- 4 t}

\int \frac{x ^{4} + 1}{x ( x ^{2} + 1 ) ^{2}} d x

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

e^{4 x}

\int_{0}^{5} \frac{1}{x ^{\frac{15}{16}}} d x

What is the integral from 1 to 4 of t^2ln(4t) ?
The integral from 1 to 4 of t^2ln(4t) is -7+254/3 ln(2)