What is the integral from-2 to 2 of 1/(x^2+4) ?

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

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integral from-2 to 2 of 1/(x^2+4)

\int_{- 2}^{2} \frac{1}{x ^{2} + 4} d x

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

\int_{- 2}^{2} \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

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

Use the common integral:

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

= \frac{1}{2} [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= \frac{1}{2} \cdot \frac{π}{2}

Simplify

= \frac{π}{4}

\frac{d}{d x} (e^{2 x + 5} + \frac{1}{e ^{x^{2}}})

\frac{\partial}{\partial x} (7 x cos (5 x y))

x y y^{^{'}} = 4 x^{2} + 2 y^{2}

\frac{d}{d x} (x \cdot y^{2})

t^{\frac{5}{2}}

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