What is the integral from 1 to 3 of 7/(16-x^2) ?

The integral from 1 to 3 of 7/(16-x^2) is (7(ln(7)-ln(5/4)+ln(3/4)))/8

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integral from 1 to 3 of 7/(16-x^2)

\int_{1}^{3} \frac{7}{16 - x ^{2}} d x

\frac{7 ( ln ( 7 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} ) )}{8}

Decimal

1.25569 \dots

Solution steps

\int_{1}^{3} \frac{7}{16 - x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int_{1}^{3} \frac{1}{16 - x ^{2}} d x

Apply Integral Substitution

= 7 \cdot \int_{\frac{1}{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

= 7 \cdot \frac{1}{4} \cdot \int_{\frac{1}{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}

= 7 \cdot \frac{1}{4} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{3}{4}}

Simplify

7 \cdot \frac{1}{4} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{3}{4}} : \frac{7}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{3}{4}}

= \frac{7}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{3}{4}}

Compute the boundaries:

\frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{2}

= \frac{7}{4} \cdot \frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{2}

Simplify

= \frac{7 ( ln ( 7 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} ) )}{8}

What is the integral from 1 to 3 of 7/(16-x^2) ?
The integral from 1 to 3 of 7/(16-x^2) is (7(ln(7)-ln(5/4)+ln(3/4)))/8