What is the integral from 1 to 3 of 1/(x^2-16) ?

The integral from 1 to 3 of 1/(x^2-16) is -(ln(7)-ln(5/4)+ln(3/4))/8

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integral from 1 to 3 of 1/(x^2-16)

\int_{1}^{3} \frac{1}{x ^{2} - 16} d x

- \frac{ln ( 7 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{8}

Decimal

- 0.17938 \dots

Solution steps

\int_{1}^{3} \frac{1}{x ^{2} - 16} d x

Apply Integral Substitution

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

Factor

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

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

Simplify

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

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

Compute the boundaries:

\frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{2}

= - \frac{1}{4} \cdot \frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{2}

Simplify

= - \frac{ln ( 7 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{8}

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What is the integral from 1 to 3 of 1/(x^2-16) ?
The integral from 1 to 3 of 1/(x^2-16) is -(ln(7)-ln(5/4)+ln(3/4))/8