What is the integral from 2 to 4 of 1/(xln^2(x)) ?

The integral from 2 to 4 of 1/(xln^2(x)) is 1/(2ln(2))

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integral from 2 to 4 of 1/(xln^2(x))

\int_{2}^{4} \frac{1}{x ln ^{2} ( x )} d x

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

Decimal Notation

0.72134 \dots

Solution steps

\int_{2}^{4} \frac{1}{x ln ^{2} ( x )} d x

Apply u-substitution

: \int_{l n (2)}^{2 l n (2)} \frac{1}{u ^{2}} d u

= \int_{l n (2)}^{2 l n (2)} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{l n (2)}^{2 l n (2)}

= [- \frac{1}{u}]_{l n (2)}^{2 l n (2)}

Compute the boundaries:

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

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

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