What is the integral from-2 to 2 of 1/(36+x^2) ?

The integral from-2 to 2 of 1/(36+x^2) is 1/3 arctan(1/3)

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integral from-2 to 2 of 1/(36+x^2)

\int_{- 2}^{2} \frac{1}{36 + x ^{2}} d x

\frac{1}{3} arctan (\frac{1}{3})

Decimal

0.10725 \dots

Solution steps

\int_{- 2}^{2} \frac{1}{36 + x ^{2}} d x

Apply Integral Substitution

= \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{6 ( 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}{6} \cdot \int_{- \frac{1}{3}}^{\frac{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}{6} [arctan (u)]_{- \frac{1}{3}}^{\frac{1}{3}}

Compute the boundaries:

2 arctan (\frac{1}{3})

= \frac{1}{6} \cdot 2 arctan (\frac{1}{3})

Simplify

= \frac{1}{3} arctan (\frac{1}{3})

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 25

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x \to 0 lim ((\frac{1}{5 x}) - (\frac{1}{e ^{5 x}} - 1))

What is the integral from-2 to 2 of 1/(36+x^2) ?
The integral from-2 to 2 of 1/(36+x^2) is 1/3 arctan(1/3)