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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

diverges

Solution steps

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

Factor

x^{2} - 1 : - (- x^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

\int_{2}^{9} (4 x + 10) d x

\int_{- 1}^{1} (\frac{12}{x ^{4}} - 3) d x

\int_{0}^{3} (1 + 6 w^{2} - 10 w^{4}) d w

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

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