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

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

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

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

e - 2

Decimal

0.71828 \dots

Solution steps

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

Apply Integration By Parts

= [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

\int \frac{x ^{2}}{x ^{3} + 1} d x

\frac{d y}{d x} = \frac{( - x e ^{- x^{2}} )}{y}

\int 8 x^{2} cos (2 x) d x

\int (x^{2} + 1)^{3} \cdot x d x

n = 1 \sum \infty (\frac{1}{n}) (\frac{1}{n})^{2}

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