What is the integral from 0 to 2 of (12)/(x^2+4) ?

The integral from 0 to 2 of (12)/(x^2+4) is (3pi)/2

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integral from 0 to 2 of (12)/(x^2+4)

\int_{0}^{2} \frac{12}{x ^{2} + 4} d x

\frac{3 π}{2}

Decimal Notation

4.71238 \dots

Solution steps

\int_{0}^{2} \frac{12}{x ^{2} + 4} d x

Take the constant out:

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

= 12 \cdot \int_{0}^{2} \frac{1}{x ^{2} + 4} d x

Apply Integral Substitution

= 12 \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

= 12 \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)

= 12 \cdot \frac{1}{2} [arctan (u)]_{0}^{1}

Simplify

12 \cdot \frac{1}{2} [arctan (u)]_{0}^{1} : 6 [arctan (u)]_{0}^{1}

= 6 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 6 \cdot \frac{π}{4}

Simplify

= \frac{3 π}{2}

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

\int 3^{3}

\int (\frac{1}{x} - 1)^{2} d x

f (x) = cos (\frac{x + 1}{x - 1})

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

What is the integral from 0 to 2 of (12)/(x^2+4) ?
The integral from 0 to 2 of (12)/(x^2+4) is (3pi)/2