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

The integral from 2 to 1 of (ln(4x))^2 is -14ln^2(2)+8ln(2)-2

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

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

- 14 ln^{2} (2) + 8 ln (2) - 2

Decimal Notation

- 3.18116 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{2} (ln (4 x))^{2} d x

Apply Integration By Parts

: [x ln^{2} (4 x) - \int 2 ln (4 x) d x]_{1}^{2}

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

\int 2 ln (4 x) d x = 2 (x ln (4 x) - x)

= - [x ln^{2} (4 x) - 2 (x ln (4 x) - x)]_{1}^{2}

Compute the boundaries:

14 ln^{2} (2) - 8 ln (2) + 2

= - (14 ln^{2} (2) - 8 ln (2) + 2)

Simplify

= - 14 ln^{2} (2) + 8 ln (2) - 2

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What is the integral from 2 to 1 of (ln(4x))^2 ?
The integral from 2 to 1 of (ln(4x))^2 is -14ln^2(2)+8ln(2)-2