What is the integral from 4 to 6 of 1/(x^2-4) ?

The integral from 4 to 6 of 1/(x^2-4) is -(ln(2)-ln(3))/4

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integral from 4 to 6 of 1/(x^2-4)

\int_{4}^{6} \frac{1}{x ^{2} - 4} d x

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

Decimal

0.10136 \dots

Solution steps

\int_{4}^{6} \frac{1}{x ^{2} - 4} d x

Apply Integral Substitution

= \int_{2}^{3} \frac{1}{2 ( u ^{2} - 1 )} d u

Take the constant out:

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

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

Factor

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

= \frac{1}{2} \cdot \int_{2}^{3} \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{2} (- \int_{2}^{3} \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

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What is the integral from 4 to 6 of 1/(x^2-4) ?
The integral from 4 to 6 of 1/(x^2-4) is -(ln(2)-ln(3))/4