What is the integral from-2 to-3 of 1/(x^2-1) ?

The integral from-2 to-3 of 1/(x^2-1) is (-ln(3)+ln(2))/2

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integral from-2 to-3 of 1/(x^2-1)

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

\frac{- ln ( 3 ) + ln ( 2 )}{2}

Decimal

- 0.20273 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Factor

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

= - \int_{- 3}^{- 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_{- 3}^{- 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}]_{- 3}^{- 2})

Simplify

- (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{- 3}^{- 2}) : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{- 3}^{- 2}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{- 3}^{- 2}

Compute the boundaries:

\frac{- ln ( 3 ) + ln ( 2 )}{2}

= \frac{- ln ( 3 ) + ln ( 2 )}{2}

\int_{- 1}^{8} - x^{2} + 7 x + 8 d x

\int_{0}^{\frac{1}{2}} sin (x^{2}) d x

\int_{- 1}^{2} (2 + ∣ x ∣) d x

\int_{0}^{2} (\frac{1}{2} x + \frac{1}{2}) - x^{2} d x

\int_{0}^{1} (\frac{1}{3 x} - 2 x + 1) d x

What is the integral from-2 to-3 of 1/(x^2-1) ?
The integral from-2 to-3 of 1/(x^2-1) is (-ln(3)+ln(2))/2