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

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

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

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

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

Decimal Notation

- 0.12770 \dots

Solution steps

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

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}

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