What is the integral from 0 to 1 of (ln(x))^2 ?

The integral from 0 to 1 of (ln(x))^2 is 2

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integral from 0 to 1 of (ln(x))^2

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

2

Solution steps

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

Apply Integration By Parts

: [x ln^{2} (x) - \int 2 ln (x) d x]_{0}^{1}

= [x ln^{2} (x) - \int 2 ln (x) d x]_{0}^{1}

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

= [x ln^{2} (x) - 2 (x ln (x) - x)]_{0}^{1}

Compute the boundaries:

2

= 2

\int_{0}^{16} \frac{x}{49 + 2 x} d x

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

\int_{0}^{1} 8 x^{2} ln (x) d x

\int_{3}^{5} x (2 - x) d x

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

What is the integral from 0 to 1 of (ln(x))^2 ?
The integral from 0 to 1 of (ln(x))^2 is 2