What is the integral from 0 to b of 1/(x^2-4) ?

The integral from 0 to b of 1/(x^2-4) is -(ln| b/2+1|-ln| b/2-1|)/4

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integral from 0 to b of 1/(x^2-4)

\int_{0}^{b} \frac{1}{x ^{2} - 4} d x

- \frac{ln \frac{b}{2} + 1 - ln \frac{b}{2} - 1}{4}

Solution steps

\int_{0}^{b} \frac{1}{x ^{2} - 4} d x

Apply Integral Substitution

= \int_{0}^{\frac{b}{2}} \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_{0}^{\frac{b}{2}} \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{2} \cdot \int_{0}^{\frac{b}{2}} \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_{0}^{\frac{b}{2}} \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}]_{0}^{\frac{b}{2}})

Simplify

\frac{1}{2} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{b}{2}}) : - \frac{1}{2} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{b}{2}}

= - \frac{1}{2} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{b}{2}}

Compute the boundaries:

\frac{ln \frac{b}{2} + 1 - ln \frac{b}{2} - 1}{2}

= - \frac{1}{2} \cdot \frac{ln \frac{b}{2} + 1 - ln \frac{b}{2} - 1}{2}

Simplify

= - \frac{ln \frac{b}{2} + 1 - ln \frac{b}{2} - 1}{4}

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What is the integral from 0 to b of 1/(x^2-4) ?
The integral from 0 to b of 1/(x^2-4) is -(ln| b/2+1|-ln| b/2-1|)/4