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

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

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

\int_{1}^{2} π (ln (x))^{2} d x

π (2 ln^{2} (2) - 4 ln (2) + 2)

Decimal Notation

0.59161 \dots

Solution steps

\int_{1}^{2} π (ln (x))^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{1}^{2} (ln (x))^{2} d x

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

2 ln^{2} (2) - 4 ln (2) + 2

= π (2 ln^{2} (2) - 4 ln (2) + 2)

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