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

The integral from 0 to 2 of 1/((x^2+4)^2) is (pi+2)/(64)

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

\int_{0}^{2} \frac{1}{( x ^{2} + 4 ) ^{2}} d x

\frac{π + 2}{64}

Decimal Notation

0.08033 \dots

Solution steps

\int_{0}^{2} \frac{1}{( x ^{2} + 4 ) ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{8 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{8} \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

= \frac{1}{8} \cdot \int_{0}^{\frac{π}{4}} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{8} \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

= \frac{1}{8} \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

= \frac{1}{8} \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}

= \frac{1}{8} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Simplify

\frac{1}{8} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : \frac{π + 2}{64}

= \frac{π + 2}{64}

\int (x^{2} - 3 x + 2) d x

x \to 0 lim (x^{2} \cdot e^{\frac{1}{x}})

\frac{d}{d x} ((x^{- 3} + 4) (x^{- 2} - 5 x))

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

What is the integral from 0 to 2 of 1/((x^2+4)^2) ?
The integral from 0 to 2 of 1/((x^2+4)^2) is (pi+2)/(64)