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

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

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

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

diverges

Solution steps

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

Simplify

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{l n (2)}

Compute the boundaries:

diverges

= diverges

\int_{1}^{\infty} \frac{1}{1 + e ^{x}} d x

2 x, 5, \frac{2}{x ^{2}}

x \to \infty lim (2 x - 4 x^{2} - 2 x)

\int \frac{1}{2 x + 2} d x

5^{t}

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