What is the integral from 5 to 10 of 1/(36-9x^2) ?

The integral from 5 to 10 of 1/(36-9x^2) is (2ln(3/2)-ln(7/2))/(36)

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integral from 5 to 10 of 1/(36-9x^2)

\int_{5}^{10} \frac{1}{36 - 9 x ^{2}} d x

\frac{2 ln ( \frac{3}{2} ) - ln ( \frac{7}{2} )}{36}

Decimal Notation

- 0.01227 \dots

Solution steps

\int_{5}^{10} \frac{1}{36 - 9 x ^{2}} d x

Apply Integral Substitution

= \int_{\frac{5}{2}}^{5} \frac{1}{18 ( - 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}{18} \cdot \int_{\frac{5}{2}}^{5} \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}{18} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{5}{2}}^{5}

Simplify

\frac{1}{18} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{5}{2}}^{5} : \frac{1}{18} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{5}{2}}^{5}

= \frac{1}{18} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{5}{2}}^{5}

Compute the boundaries:

\frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( \frac{7}{2} ) + ln ( \frac{3}{2} )}{2}

= \frac{1}{18} \cdot \frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( \frac{7}{2} ) + ln ( \frac{3}{2} )}{2}

Simplify

= \frac{2 ln ( \frac{3}{2} ) - ln ( \frac{7}{2} )}{36}

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What is the integral from 5 to 10 of 1/(36-9x^2) ?
The integral from 5 to 10 of 1/(36-9x^2) is (2ln(3/2)-ln(7/2))/(36)