Du inside arcsin

$\displaystyle \int_{0}^{\infty} \dfrac{\arcsin(\dfrac{2x}{1+x^{2}})}{1+x^2}dx=\int_{0}^{1} \dfrac{\arcsin(\dfrac{2x}{1+x^{2}})}{1+x^2}dx+\int_{1}^{\infty} \dfrac{\arcsin(\dfrac{2x}{1+x^{2}})}{1+x^2}dx$

Then, $\displaystyle \int_{1}^{\infty} \dfrac{\arcsin(\dfrac{2x}{1+x^{2}})}{1+x^2}dx=\int_{1}^{0} \dfrac{\arcsin(\dfrac{2u}{1+u^{2}})}{1+u^2}-du=\int_{0}^{1} \dfrac{\arcsin(\dfrac{2u}{1+u^{2}})}{1+u^2}du$ by change $x=\dfrac{1}{u}$

$\displaystyle \int_{0}^{\infty} \dfrac{\arcsin(\dfrac{2u}{1+u^{2}})}{1+u^2}du=2 \left ( \int_{0}^{1} \dfrac{\arcsin(\dfrac{2u}{1+u^{2}})}{1+u^2}du \right )$

Now, change $\displaystyle u=\tan(\dfrac{y}{2})$ . Thus the integral become

$\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{y}{2} dy= \dfrac{y^{2}}{4}=\dfrac{\pi^{2}}{16}$

$\displaystyle \int_{0}^{\infty} \dfrac{\arcsin(\dfrac{2u}{1+u^{2}})}{1+u^2}du=2 \left ( \int_{0}^{1} \dfrac{\arcsin(\dfrac{2u}{1+u^{2}})}{1+u^2}du \right )$ $=2\times \dfrac{\pi^{2}}{16}=\boxed{\dfrac{\pi^{2}}{8}}$

$\begin{aligned} I & = \int_0^\infty \frac {\arcsin \left(\frac {2x}{1+x^2}\right)}{1+x^2}\ dx & \small \color{#3D99F6} \text{Let }x = \tan \frac \theta 2 \implies dx = \frac 12 \sec^2 \frac \theta 2 \ d\theta \\ & = \int_0^\pi \frac {\arcsin \left(\sin \theta \right)}2\ d\theta & \small \color{#3D99F6} \text{then } \sin \theta = \frac {2x}{1+x^2} \\ & = \frac 12 \left(\int_0^\frac \pi 2 \arcsin (\sin \theta) \ d\theta + \int_\frac \pi 2^\pi \arcsin (\sin \theta) \ d\theta \right) & \small \color{#3D99F6} \text{Note that }- \frac \pi 2 \le \arcsin x \le \frac \pi 2 \\ & = \frac 12 \left(\int_0^\frac \pi 2 \theta \ d\theta + \int_\frac \pi 2^\pi (\pi - \theta) \ d\theta \right) \\ & = \frac 12 \left(\frac {\theta^2}2 \bigg|_0^\frac \pi 2 + \left[\pi \theta - \frac {\theta^2}2\right]_\frac \pi 2^\pi \right) \\ & = \frac 12 \left(\frac {\pi^2}8 - 0 + \pi^2 - \frac {\pi^2}2 - \frac {\pi^2}2 + \frac {\pi^2}8 \right) \\ & = \frac {\pi^2}8 \end{aligned}$

Therefore, $a+b = 2+8 = \boxed{10}$ .

First multiply by 2 in the integrand and 1/2 outside of it is the same thing as multiplying by 1. Recognizing that 2x/1+x^2 and 2/1+x^2 are just reversed versions of sin(x) and dx with regard to the Tangent-Half Angle Substitution allows us to perform the so-called "Reverse Weierstrass" substitution. So we end up having 1/2 times the integral of u du (Note: Since the bounds are in the domain of the arcsine function, then arcsin(sin(u)) = u), from 0 to pi/2 which equals pi^2/(4*2) = pi^2 / 8. Then 2 + 8 = 10, as desired.