What is the integral from 0 to 9 of 1/(81+x^2) ?

The integral from 0 to 9 of 1/(81+x^2) is pi/(36)

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integral from 0 to 9 of 1/(81+x^2)

\int_{0}^{9} \frac{1}{81 + x ^{2}} d x

\frac{π}{36}

Decimal Notation

0.08726 \dots

Solution steps

\int_{0}^{9} \frac{1}{81 + x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{9 ( 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}{9} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{9} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{9} \cdot \frac{π}{4}

Simplify

= \frac{π}{36}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 20 e^{t}, y (0) = - 4, y^{^{'}} (0) = 1

\int_{1}^{4} 6 e^{x} d x

f (x, y) = (x)^{2} + (y)^{4}

x \to 3 + lim (\frac{1}{∣ x - 3 ∣})

y \to 0 lim (\frac{y ^{2}}{y ^{4}})

What is the integral from 0 to 9 of 1/(81+x^2) ?
The integral from 0 to 9 of 1/(81+x^2) is pi/(36)