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

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

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{3} \frac{1}{2 u} d u

= \int_{0}^{3} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{0}^{3}

Compute the boundaries:

diverges

= diverges

\int_{0}^{1} 2 + x^{2} d x

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\int_{0}^{3} f (t) d t

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