What is the integral from 0 to 3 of 1/(9-x^2) ?

The integral from 0 to 3 of 1/(9-x^2) is diverges

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integral from 0 to 3 of 1/(9-x^2)

\int_{0}^{3} \frac{1}{9 - x ^{2}} d x

diverges

Solution steps

\int_{0}^{3} \frac{1}{9 - x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \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}^{1} \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}^{1}

Simplify

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

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

Compute the boundaries:

diverges

= diverges

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\int_{\frac{1}{2}}^{2} ∣ ln (x) ∣ d x

What is the integral from 0 to 3 of 1/(9-x^2) ?
The integral from 0 to 3 of 1/(9-x^2) is diverges