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

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

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

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

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

Decimal Notation

1.73286 \dots

Solution steps

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

Factor

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

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

Simplify

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

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

Compute the boundaries:

\frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 3 )}{2}

= - 5 \cdot \frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 3 )}{2}

Simplify

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

x \to 5 - lim (\frac{x ^{5}}{( x - 5 )})

\frac{2 x ^{2} + 8 x + 2}{x}

\int \frac{cos ( x ) - tan ( x )}{cos ^{2} ( x )} d x

4 x

\frac{d}{d x} (sin (x) + 4 x^{2})

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