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

The integral from 2 to 4 of 1/(x^2-3) is -(ln(\frac{37-20sqrt(3))/(13))}{2sqrt(3)}

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

\int_{2}^{4} \frac{1}{x ^{2} - 3} d x

- \frac{ln ( \frac{37 - 20 3}{13} )}{2 3}

Solution steps

\int_{2}^{4} \frac{1}{x ^{2} - 3} d x

Apply Integral Substitution

= \int_{\frac{2}{3}}^{\frac{4}{3}} \frac{1}{3 ( 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}{3} \cdot \int_{\frac{2}{3}}^{\frac{4}{3}} \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{3} \cdot \int_{\frac{2}{3}}^{\frac{4}{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}{3} (- \int_{\frac{2}{3}}^{\frac{4}{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}{3} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{2}{3}}^{\frac{4}{3}})

Simplify

\frac{1}{3} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{2}{3}}^{\frac{4}{3}}) : - \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{2}{3}}^{\frac{4}{3}}

= - \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{2}{3}}^{\frac{4}{3}}

Compute the boundaries:

\frac{ln ( \frac{4}{3} + 1 ) - ln ( \frac{4}{3} - 1 ) - ln ( \frac{2}{3} + 1 ) + ln ( \frac{2}{3} - 1 )}{2}

= - \frac{1}{3} \cdot \frac{ln ( \frac{4}{3} + 1 ) - ln ( \frac{4}{3} - 1 ) - ln ( \frac{2}{3} + 1 ) + ln ( \frac{2}{3} - 1 )}{2}

Simplify

= - \frac{ln ( \frac{37 - 20 3}{13} )}{2 3}

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What is the integral from 2 to 4 of 1/(x^2-3) ?
The integral from 2 to 4 of 1/(x^2-3) is -(ln(\frac{37-20sqrt(3))/(13))}{2sqrt(3)}