What is the integral from 0 to 2 of (1/((9+x^2)^2)) ?

The integral from 0 to 2 of (1/((9+x^2)^2)) is 1/54 arctan(2/3)+1/117

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integral from 0 to 2 of (1/((9+x^2)^2))

\int_{0}^{2} (\frac{1}{( 9 + x ^{2} ) ^{2}}) d x

\frac{1}{54} arctan (\frac{2}{3}) + \frac{1}{117}

Decimal

0.01943 \dots

Solution steps

\int_{0}^{2} \frac{1}{( 9 + x ^{2} ) ^{2}} d x

Apply Integral Substitution

= \int_{0}^{\frac{2}{3}} \frac{1}{27 ( 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}{27} \cdot \int_{0}^{\frac{2}{3}} \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{2}{3})} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= \frac{1}{27} \cdot \int_{0}^{a r c t a n (\frac{2}{3})} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{27} \cdot \int_{0}^{a r c t a n (\frac{2}{3})} \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}{27} \cdot \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{3})} 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}{27} \cdot \frac{1}{2} (\int_{0}^{a r c t a n (\frac{2}{3})} 1 d v + \int_{0}^{a r c t a n (\frac{2}{3})} cos (2 v) d v)

\int_{0}^{a r c t a n (\frac{2}{3})} 1 d v = arctan (\frac{2}{3})

\int_{0}^{a r c t a n (\frac{2}{3})} cos (2 v) d v = \frac{6}{13}

= \frac{1}{27} \cdot \frac{1}{2} (arctan (\frac{2}{3}) + \frac{6}{13})

Simplify

\frac{1}{27} \cdot \frac{1}{2} (arctan (\frac{2}{3}) + \frac{6}{13}) : \frac{1}{54} arctan (\frac{2}{3}) + \frac{1}{117}

= \frac{1}{54} arctan (\frac{2}{3}) + \frac{1}{117}

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What is the integral from 0 to 2 of (1/((9+x^2)^2)) ?
The integral from 0 to 2 of (1/((9+x^2)^2)) is 1/54 arctan(2/3)+1/117