What is the integral from sqrt(3) to 3 of 1/(9+x^2) ?

The integral from sqrt(3) to 3 of 1/(9+x^2) is pi/(36)

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integral from sqrt(3) to 3 of 1/(9+x^2)

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

\frac{π}{36}

Decimal Notation

0.08726 \dots

Solution steps

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

Apply Integral Substitution

= \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

= \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)

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

Compute the boundaries:

\frac{π}{12}

= \frac{1}{3} \cdot \frac{π}{12}

Simplify

= \frac{π}{36}

\int_{- \infty}^{0} \frac{1}{3 - 9 x} d x

\frac{d}{d x} ((f (x))^{2})

\frac{d}{d y} (1 - 2 y)

\int \frac{ln ( x ^{2} )}{2 x} d x

y = 3 x^{2} + 14 x - 5

What is the integral from sqrt(3) to 3 of 1/(9+x^2) ?
The integral from sqrt(3) to 3 of 1/(9+x^2) is pi/(36)