What is the integral from-1 to 2 of 1/(x^2-9) ?

The integral from-1 to 2 of 1/(x^2-9) is -(ln(10))/6

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integral from-1 to 2 of 1/(x^2-9)

\int_{- 1}^{2} \frac{1}{x ^{2} - 9} d x

- \frac{ln ( 10 )}{6}

Decimal

- 0.38376 \dots

Solution steps

\int_{- 1}^{2} \frac{1}{x ^{2} - 9} d x

Apply Integral Substitution

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

Factor

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

= - \frac{ln ( 10 )}{6}

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\int_{0}^{a} \frac{1}{a ^{2} - x ^{2}} d x

What is the integral from-1 to 2 of 1/(x^2-9) ?
The integral from-1 to 2 of 1/(x^2-9) is -(ln(10))/6