What is the integral from 3 to 4 of 1/(25-x^2) ?

The integral from 3 to 4 of 1/(25-x^2) is (ln(9/4))/(10)

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integral from 3 to 4 of 1/(25-x^2)

\int_{3}^{4} \frac{1}{25 - x ^{2}} d x

\frac{ln ( \frac{9}{4} )}{10}

Decimal Notation

0.08109 \dots

Solution steps

\int_{3}^{4} \frac{1}{25 - x ^{2}} d x

Apply Integral Substitution

= \int_{\frac{3}{5}}^{\frac{4}{5}} \frac{1}{5 ( - 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}{5} \cdot \int_{\frac{3}{5}}^{\frac{4}{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}{5} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{3}{5}}^{\frac{4}{5}}

Simplify

\frac{1}{5} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{3}{5}}^{\frac{4}{5}} : \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{3}{5}}^{\frac{4}{5}}

= \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{3}{5}}^{\frac{4}{5}}

Compute the boundaries:

\frac{ln ( \frac{9}{5} ) + ln ( 5 ) - ln ( \frac{8}{5} ) + ln ( \frac{2}{5} )}{2}

= \frac{1}{5} \cdot \frac{ln ( \frac{9}{5} ) + ln ( 5 ) - ln ( \frac{8}{5} ) + ln ( \frac{2}{5} )}{2}

Simplify

= \frac{ln ( \frac{9}{4} )}{10}

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What is the integral from 3 to 4 of 1/(25-x^2) ?
The integral from 3 to 4 of 1/(25-x^2) is (ln(9/4))/(10)