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

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

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

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

- \frac{ln ( \frac{3}{2} ) - ln ( 5 ) + ln ( 3 )}{4}

Decimal

0.02634 \dots

Solution steps

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

\frac{1}{2 x ^{2} - 2} = \frac{1}{2} \frac{1}{x ^{2} - 1}

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

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

Factor

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

= \frac{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

= \frac{1}{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}

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

Simplify

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

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

Compute the boundaries:

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

= - \frac{1}{2} \cdot \frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

Simplify

= - \frac{ln ( \frac{3}{2} ) - ln ( 5 ) + ln ( 3 )}{4}

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