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

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

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

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

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

Decimal Notation

0.44269 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

\int_{- 1}^{1} sec^{2} (x) d x

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

\int_{0}^{2} x^{2} - 3 x d x

\int_{0}^{4} (2 x + 1)^{- \frac{1}{2}} d x

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

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