What is the integral from-1 to 1 of 1/(16-9x^2) ?

The integral from-1 to 1 of 1/(16-9x^2) is 1/12 ln(7)

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integral from-1 to 1 of 1/(16-9x^2)

\int_{- 1}^{1} \frac{1}{16 - 9 x ^{2}} d x

\frac{1}{12} ln (7)

Decimal

0.16215 \dots

Solution steps

\int_{- 1}^{1} \frac{1}{16 - 9 x ^{2}} d x

Apply Integral Substitution

= \int_{- \frac{3}{4}}^{\frac{3}{4}} \frac{1}{12 ( - 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}{12} \cdot \int_{- \frac{3}{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}{12} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{3}{4}}^{\frac{3}{4}}

Simplify

\frac{1}{12} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{3}{4}}^{\frac{3}{4}} : \frac{1}{12} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{3}{4}}^{\frac{3}{4}}

= \frac{1}{12} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{3}{4}}^{\frac{3}{4}}

Compute the boundaries:

\frac{4 ln ( 2 ) + 2 ln ( \frac{7}{4} )}{2}

= \frac{1}{12} \cdot \frac{4 ln ( 2 ) + 2 ln ( \frac{7}{4} )}{2}

Simplify

= \frac{1}{12} ln (7)

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What is the integral from-1 to 1 of 1/(16-9x^2) ?
The integral from-1 to 1 of 1/(16-9x^2) is 1/12 ln(7)