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

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

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

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

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

Decimal Notation

- 0.54930 \dots

Solution steps

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

Apply Integral Substitution

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

Factor

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

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