What is the integral from 4 to 5 of 2/(x^2-1) ?

The integral from 4 to 5 of 2/(x^2-1) is -ln(3/2)+ln(5)-ln(3)

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integral from 4 to 5 of 2/(x^2-1)

\int_{4}^{5} \frac{2}{x ^{2} - 1} d x

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

Decimal Notation

0.10536 \dots

Solution steps

\int_{4}^{5} \frac{2}{x ^{2} - 1} d x

Take the constant out:

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

= 2 \cdot \int_{4}^{5} \frac{1}{x ^{2} - 1} d x

Factor

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

= 2 \cdot \int_{4}^{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

= 2 (- \int_{4}^{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}

= 2 (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{5})

Simplify

2 (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{5}) : - 2 [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{5}

= - 2 [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{5}

Compute the boundaries:

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

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

Simplify

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

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

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\int_{1}^{5} 3 x^{4} sin (2 x) d x

\int_{0}^{\infty} \frac{1}{e ^{x} + 3} d x

What is the integral from 4 to 5 of 2/(x^2-1) ?
The integral from 4 to 5 of 2/(x^2-1) is -ln(3/2)+ln(5)-ln(3)