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

The integral from e to 2e of 1/(xln^2(x)) is -1/(ln(2e))+1

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

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

- \frac{1}{ln ( 2 e )} + 1

Decimal Notation

0.40938 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{ln ( 2 e )} + 1

= - \frac{1}{ln ( 2 e )} + 1

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