What is the integral from sqrt(3) to 3 of 8/(9+x^2) ?

The integral from sqrt(3) to 3 of 8/(9+x^2) is (2pi)/9

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integral from sqrt(3) to 3 of 8/(9+x^2)

\int_{3}^{3} \frac{8}{9 + x ^{2}} d x

\frac{2 π}{9}

Decimal

0.69813 \dots

Solution steps

\int_{3}^{3} \frac{8}{9 + x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int_{3}^{3} \frac{1}{9 + x ^{2}} d x

Apply Integral Substitution

= 8 \cdot \int_{\frac{1}{3}}^{1} \frac{1}{3 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{3} \cdot \int_{\frac{1}{3}}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 8 \cdot \frac{1}{3} [arctan (u)]_{\frac{1}{3}}^{1}

Simplify

8 \cdot \frac{1}{3} [arctan (u)]_{\frac{1}{3}}^{1} : \frac{8}{3} [arctan (u)]_{\frac{1}{3}}^{1}

= \frac{8}{3} [arctan (u)]_{\frac{1}{3}}^{1}

Compute the boundaries:

\frac{π}{12}

= \frac{8}{3} \cdot \frac{π}{12}

Simplify

= \frac{2 π}{9}

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What is the integral from sqrt(3) to 3 of 8/(9+x^2) ?
The integral from sqrt(3) to 3 of 8/(9+x^2) is (2pi)/9