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

The integral from 0 to 2 of pi(4/(x^2+4))^2 is pi(pi/4+1/2)

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

\int_{0}^{2} π (\frac{4}{x ^{2} + 4})^{2} d x

π (\frac{π}{4} + \frac{1}{2})

Decimal Notation

4.03819 \dots

Solution steps

\int_{0}^{2} π (\frac{4}{x ^{2} + 4})^{2} d x

Take the constant out:

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

= π \cdot \int_{0}^{2} (\frac{4}{x ^{2} + 4})^{2} d x

Apply Integral Substitution

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= π 2 \cdot \int_{0}^{\frac{π}{4}} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= π 2 \cdot \int_{0}^{\frac{π}{4}} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= π 2 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π 2 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d v + \int_{0}^{\frac{π}{4}} cos (2 v) d v)

\int_{0}^{\frac{π}{4}} 1 d v = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 v) d v = \frac{1}{2}

= π 2 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Simplify

π 2 \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : π (\frac{π}{4} + \frac{1}{2})

= π (\frac{π}{4} + \frac{1}{2})

x \to 0 lim (\frac{( sin ( x ) )}{x})

\frac{\partial}{\partial x} (- x^{3} + 6 x^{2} y^{3} + 5 y^{2})

\int \frac{2 x + 3}{( 2 x ^{2} + 6 x ) ^{4}} d x

\frac{d x}{d t} = \frac{1}{t ( 1 - t )}

\frac{d}{d t} (x y)

What is the integral from 0 to 2 of pi(4/(x^2+4))^2 ?
The integral from 0 to 2 of pi(4/(x^2+4))^2 is pi(pi/4+1/2)