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

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

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

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

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

Decimal Notation

0.18831 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

y = - x, x = - 1, x = 2, y = 0

y^{^{''}} + 8 y^{^{'}} + 65 y = 455, y (0) = 0, y^{^{'}} (0) = 0

\int \frac{1}{x ^{2} + 6 x + 18} d x

x \to 1 lim (x - 4)

\int \frac{( arctan ( x ) ) ^{5}}{x ^{2} + 1} d x

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