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

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

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

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

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

Decimal Notation

0.12770 \dots

Solution steps

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

Apply Integral Substitution

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

Factor

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

= \frac{1}{2} \cdot \int_{\frac{3}{2}}^{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{3}{2}}^{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{3}{2}}^{2})

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

f (x) = 5 x^{2} - 3 x + 7

\int \frac{1}{4 + 2 sin ( x )} d x

e^{- 5 x^{2}}

x \to - 4 lim (15)

n = 1 \sum \infty (\frac{1}{2})^{2}

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