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

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

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{l n (2)} \frac{1}{u ^{2}} d u

= \int_{0}^{l n (2)} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{0}^{l n (2)}

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

Compute the boundaries:

diverges

= diverges

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

\int_{0}^{1} 3 x^{2} e^{x} d x

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