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

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

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

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

e - 2

Decimal Notation

0.71828 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

e - 2

= e - 2

x \to 5 lim (\frac{5}{x - 5})

\int \frac{1}{8} x d x

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

(3, 2), (6, 8)

3 x + 2 cos (x)

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