What is the integral from 0 to 1 of 1/(x^2-9) ?

The integral from 0 to 1 of 1/(x^2-9) is -(ln(2))/6

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integral from 0 to 1 of 1/(x^2-9)

\int_{0}^{1} \frac{1}{x ^{2} - 9} d x

- \frac{ln ( 2 )}{6}

Decimal Notation

- 0.11552 \dots

Solution steps

\int_{0}^{1} \frac{1}{x ^{2} - 9} d x

Apply Integral Substitution

= \int_{0}^{\frac{1}{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_{0}^{\frac{1}{3}} \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{3} \cdot \int_{0}^{\frac{1}{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_{0}^{\frac{1}{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}]_{0}^{\frac{1}{3}})

Simplify

\frac{1}{3} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{3}}) : - \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{3}}

= - \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{3}}

Compute the boundaries:

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

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

Simplify

= - \frac{ln ( 2 )}{6}

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What is the integral from 0 to 1 of 1/(x^2-9) ?
The integral from 0 to 1 of 1/(x^2-9) is -(ln(2))/6