What is the integral from 0 to 2 of 3/(x^2+4) ?

The integral from 0 to 2 of 3/(x^2+4) is (3pi)/8

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integral from 0 to 2 of 3/(x^2+4)

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

\frac{3 π}{8}

Decimal Notation

1.17809 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{0}^{2} \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

= 3 \cdot \int_{0}^{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

= 3 \cdot \frac{1}{2} \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)

= 3 \cdot \frac{1}{2} [arctan (u)]_{0}^{1}

Simplify

3 \cdot \frac{1}{2} [arctan (u)]_{0}^{1} : \frac{3}{2} [arctan (u)]_{0}^{1}

= \frac{3}{2} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{3}{2} \cdot \frac{π}{4}

Simplify

= \frac{3 π}{8}

\frac{d}{d x} (arccos (2 x))

\frac{\partial}{\partial x} (12 x y e^{- x^{2} - y^{2}})

x \to 0 lim (3 x + 41)

\frac{\partial}{\partial y} (2 x^{3} + 4 y - 2 z^{2})

\frac{d}{d x} (\frac{x ^{2} - 3 x - 54}{x + 6})

What is the integral from 0 to 2 of 3/(x^2+4) ?
The integral from 0 to 2 of 3/(x^2+4) is (3pi)/8