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

The integral from 0 to 2 of (1/(1+x^2))^2 is 1/2 arctan(2)+1/5

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

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

\frac{1}{2} arctan (2) + \frac{1}{5}

Decimal

0.75357 \dots

Solution steps

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

Simplify

(\frac{1}{1 + x ^{2}})^{2} : \frac{1}{( 1 + x ^{2} ) ^{2}}

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

Apply Trigonometric Substitution

= \int_{0}^{a r c t a n (2)} \frac{1}{sec ^{4} ( u ) cos ^{2} ( u )} d u

Rewrite using trig identities

= \int_{0}^{a r c t a n (2)} \frac{1 + cos ( 2 u )}{2} 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_{0}^{a r c t a n (2)} 1 + cos (2 u) d u

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}{2} (\int_{0}^{a r c t a n (2)} 1 d u + \int_{0}^{a r c t a n (2)} cos (2 u) d u)

\int_{0}^{a r c t a n (2)} 1 d u = arctan (2)

\int_{0}^{a r c t a n (2)} cos (2 u) d u = \frac{2}{5}

= \frac{1}{2} (arctan (2) + \frac{2}{5})

Simplify

\frac{1}{2} (arctan (2) + \frac{2}{5}) : \frac{1}{2} arctan (2) + \frac{1}{5}

= \frac{1}{2} arctan (2) + \frac{1}{5}

What is the integral from 0 to 2 of (1/(1+x^2))^2 ?
The integral from 0 to 2 of (1/(1+x^2))^2 is 1/2 arctan(2)+1/5