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

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

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

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

diverges

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2 ln ( x ) - ln ^{2} ( x )}{x} d x = ln^{2} (x) - \frac{ln ^{3} ( x )}{3}

= [ln^{2} (x) - (ln^{2} (x) - \frac{ln ^{3} ( x )}{3})]_{0}^{1}

Simplify

[ln^{2} (x) - (ln^{2} (x) - \frac{ln ^{3} ( x )}{3})]_{0}^{1} : [\frac{1}{3} ln^{3} (x)]_{0}^{1}

= [\frac{1}{3} ln^{3} (x)]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{1}^{2} 2 x + 5 d x

\int_{1}^{6} 2 x + \frac{1}{x} d x

\int_{- \frac{π}{6}}^{\frac{π}{6}} 2 sec^{2} (x) d x

\int_{- 1}^{3} x^{2} + 1 d x

\int_{2}^{4} \frac{1}{x - 1} d x

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