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

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

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

\int_{1}^{e^{2}} π (ln (x))^{2} d x

π (2 e^{2} - 2)

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

2 e^{2} - 2

= π (2 e^{2} - 2)

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