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

Simplify

(- ln (x))^{2} : ln^{2} (x)

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

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\int_{0}^{1} (1 - 8 x^{3} + 16 x^{7}) d x

\int_{1}^{2} (10 x^{3} + \frac{6}{x ^{2}}) 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