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

The integral from 3 to 9 of 1/(9+x^2) is 1/3 (arctan(3)-pi/4)

Popular >

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

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

\frac{1}{3} (arctan (3) - \frac{π}{4})

Decimal

0.15454 \dots

Solution steps

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

Apply Integral Substitution

= \int_{1}^{3} \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_{1}^{3} \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)]_{1}^{3}

Compute the boundaries:

arctan (3) - \frac{π}{4}

= \frac{1}{3} (arctan (3) - \frac{π}{4})

\int_{0}^{4} 9 + 2 x d x

\int_{0}^{1} sin^{2} (kπ x) d x

\int_{0}^{4} \frac{2}{3 x + 4} d x

\int_{- 1}^{2} - x^{2} + 6 d x

\int_{2}^{7} t^{5} ln (t) d t

What is the integral from 3 to 9 of 1/(9+x^2) ?
The integral from 3 to 9 of 1/(9+x^2) is 1/3 (arctan(3)-pi/4)