What is the integral from 2 to 3 of 5/(x^2-1) ?

The integral from 2 to 3 of 5/(x^2-1) is -(5(ln(2)-ln(3)))/2

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integral from 2 to 3 of 5/(x^2-1)

\int_{2}^{3} \frac{5}{x ^{2} - 1} d x

- \frac{5 ( ln ( 2 ) - ln ( 3 ) )}{2}

Decimal

1.01366 \dots

Solution steps

\int_{2}^{3} \frac{5}{x ^{2} - 1} d x

Take the constant out:

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

= 5 \cdot \int_{2}^{3} \frac{1}{x ^{2} - 1} d x

Factor

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

= 5 \cdot \int_{2}^{3} \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

= 5 (- \int_{2}^{3} \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}

= 5 (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{2}^{3})

Simplify

5 (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{2}^{3}) : - 5 [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{2}^{3}

= - 5 [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{2}^{3}

Compute the boundaries:

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

= - 5 \cdot \frac{ln ( 2 ) - ln ( 3 )}{2}

Simplify

= - \frac{5 ( ln ( 2 ) - ln ( 3 ) )}{2}

\int \frac{1}{e ^{x}} d x

\frac{d x}{d t} = 2 x

x \to \infty lim 1 + \frac{1}{x} - 1 x

y^{^{'}} = e^{x - y}

\int_{- 2}^{2} 5 - 4 - x^{2} d x

What is the integral from 2 to 3 of 5/(x^2-1) ?
The integral from 2 to 3 of 5/(x^2-1) is -(5(ln(2)-ln(3)))/2