An inspired problem (but I shouldn't say what inspired me)

Let $u = \dfrac{2x}{1+x^2}$ . Then $\dfrac{du}{dx} = \dfrac{2(1+x^2) - (2x)^2}{(1+x^2)^2}$ .

The integral becomes $\displaystyle \int_{0}^{\sqrt{2}/2} \dfrac{du}{\sqrt{1-u^2}}$

$= \left. \arcsin u \right|_{0}^{\sqrt{2}/2} = \dfrac{\pi}{4} \to 1+4 = \boxed{5}$

$\large = \displaystyle {\int \limits_{0}^{\sqrt{3}}} \frac{2+2x^2 - 4x^2}{(1+x^2)^2 \sqrt{\left(1 + \dfrac{2x}{1+x^2} \right) \left(1 - \dfrac{2x}{1+x^2} \right) }} \, dx$ $\large = \displaystyle {\int \limits_{0}^{\sqrt{3}}} \frac{2(1- x^2)}{(1+x^2)^2 \sqrt{\left(\dfrac{x^2 +2x +1}{1+x^2} \right) \left(\dfrac{x^2 -2x +1}{1+x^2} \right) }} \, dx$ $\large = \displaystyle {\int \limits_{0}^{\sqrt{3}}} \frac{2(1+x)(1-x)}{(1+x^2)^2 \sqrt{\left(\dfrac{(x+1)^2 (x-1)^2}{(1+x^2)^2} \right) }} \, dx$ $\large = \displaystyle {\int \limits_{0}^{\sqrt{3}}} \frac{2(1+x)(1-x)}{(1+x^2)^2 \dfrac{(x+1)(x-1)}{(1+x^2)} } \, dx$ Simplified: $\large \displaystyle {\int \limits_{0}^{\sqrt{3}}} \frac{2}{1+x^2} \, dx$ $\large \displaystyle {2 \int \limits_{0}^{\sqrt{3}}} \frac{1}{1+x^2} \, dx$ Note that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$ . Applying this, we get $\large \displaystyle {2 \left( \tan^{-1} \left( \sqrt{3} \right) - \tan^{-1} 0 \right) }$ $\large \displaystyle {2 \left( \frac{\pi}{3} - 0 \right) }$ $\large \displaystyle {\frac{2\pi}{3} }$ Final answer: $\large 2+3=\boxed{5}$

Let x = tany

Hence changing limits from 0 to 3/8 pi

Our substitution is valid as function is one one and continuous

Simplify a bit you will get integral as 2cos2y / [cos2y]

where [.] denotes absolute value function

Break the integral into parts first from 0 to 1/4 pi and second from 1/4 pi to 3/8 pi as cos2y will be positive in one and negative in other

Solving you will get required answer

Just simplifying the algebraic expression leads to 2/1+x^2 . No need to sub anything