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

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

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

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

\frac{1}{ln ( 2 )}

Decimal

1.44269 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1 - ln ^{2} ( x ) \cdot \int - \frac{l n ( x ) + 2}{x l n ^{3} ( x )} d x}{ln ^{2} ( x )}]_{0}^{\frac{1}{2}}

= [\frac{1 - ln ^{2} ( x ) \cdot \int - \frac{l n ( x ) + 2}{x l n ^{3} ( x )} d x}{ln ^{2} ( x )}]_{0}^{\frac{1}{2}}

\int - \frac{ln ( x ) + 2}{x ln ^{3} ( x )} d x = \frac{1}{ln ( x )} + \frac{1}{ln ^{2} ( x )}

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

\frac{1 - ln ^{2} ( x ) ( \frac{1}{l n ( x )} + \frac{1}{l n ^{2} ( x )} )}{ln ^{2} ( x )} = - \frac{1}{ln ( x )}

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

Compute the boundaries:

\frac{1}{ln ( 2 )}

= \frac{1}{ln ( 2 )}

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