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

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

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

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

diverges

Solution steps

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

Use the common integral:

\int \frac{1}{1 - x ^{2}} d x = \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}

= [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{1}^{2}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{1}^{2} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{1}^{2}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{1}^{2}

Compute the boundaries:

diverges

= diverges

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

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

\int_{1}^{2} \frac{x + 3}{x ( x + 2 )} d x

\int_{1}^{5} x^{2} + 2 d x

\int_{0}^{3} (4 t - 1)^{2} d t

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