What is the integral from e to 3 of 1/(16-x^2) ?

The integral from e to 3 of 1/(16-x^2) is (ln(7)-ln(e/4+1)+ln(-e/4+1))/8

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integral from e to 3 of 1/(16-x^2)

\int_{e}^{3} \frac{1}{16 - x ^{2}} d x

\frac{ln ( 7 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{8}

Decimal

0.03615 \dots

Solution steps

\int_{e}^{3} \frac{1}{16 - x ^{2}} d x

Apply Integral Substitution

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

Simplify

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

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

Compute the boundaries:

\frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{2}

= \frac{1}{4} \cdot \frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{2}

Simplify

= \frac{ln ( 7 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{8}

\frac{d}{d x} (5 x - 2)

\int_{0}^{π} 1 + sin (x) d x

\frac{d y}{d t} = t \cdot y

\int_{0}^{1} ln (1 + x) d x

\frac{d}{d x} (\frac{10 x + 4 x ^{2}}{( 5 + 4 x ) ^{2}})

What is the integral from e to 3 of 1/(16-x^2) ?
The integral from e to 3 of 1/(16-x^2) is (ln(7)-ln(e/4+1)+ln(-e/4+1))/8