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

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

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

\int_{1}^{e} π (ln (x))^{2} d x

π (e - 2)

Decimal

2.25654 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

e - 2

= π (e - 2)

\int_{- 1}^{1} 1^{2} - x^{2} d x

\int_{0}^{1} (2 x - 1) d x

\int_{1}^{3} \frac{18 ( ln ( x ) ) ^{5}}{x} d x

\int_{- 2}^{0} \frac{13}{6} x^{3} d x

\int_{- 2}^{- 1} \frac{x}{( x ^{2} + 2 ) ^{3}} d x

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