What is the integral from 1 to 4 of 1/(x^2-1) ?

The integral from 1 to 4 of 1/(x^2-1) is diverges

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integral from 1 to 4 of 1/(x^2-1)

\int_{1}^{4} \frac{1}{x ^{2} - 1} d x

diverges

Solution steps

\int_{1}^{4} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{1}^{4} \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}^{4} \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}^{4}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{1}^{4} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{1}^{4}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{1}^{4}

Compute the boundaries:

diverges

= diverges

\int_{- 1}^{3} 2 x + 6 d x

\int_{0}^{1} (x - x e^{x - 1}) d x

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

\int_{- 1}^{3} 2 x + 4 d x

\int_{0}^{3} (y - 3) (2 y + 3) d y

What is the integral from 1 to 4 of 1/(x^2-1) ?
The integral from 1 to 4 of 1/(x^2-1) is diverges