Inverse is Integral in Maths!

Relevant wiki: Inverse Trigonometric Functions

$\displaystyle a = \int_0^{1} \arctan (1- x + x^2) \, dx$

$\displaystyle a = \int_0^{1} \frac{\pi}{2} - arccot (1- x + x^2) \, dx$

$\displaystyle a = \int_0^{1} \frac{\pi}{2} - \arctan (\frac{1}{1- x + x^2}) \, dx$

$\displaystyle a = \int_0^{1} \frac{\pi}{2} - \arctan (\frac{x-(x-1)}{1+ x(x-1)}) \, dx$

$\displaystyle a = \int_0^{1} \frac{\pi}{2} - \arctan \tan (\arctan x - \arctan (x-1)) \, dx$

$\displaystyle a = \int_0^{1} \frac{\pi}{2} - \arctan x + \arctan (x-1) \, dx$

$\displaystyle a = \frac{\pi}{2} - \int_0^{1} 1 \cdot \arctan x + \int_0^{1} 1 \cdot \arctan (x-1) \, dx$

Now integrate the 2 terms by parts to get,

$\displaystyle a = \frac{\pi}{2} - [ x \arctan x -\frac{1}{2} \ln (1+ x^2)]^1_0 + [ (x-1) \arctan (x-1) -\frac{1}{2} \ln (1+ (x-1)^2)]^1_0$

$\displaystyle a = \frac{\pi}{2} - [ \frac{\pi}{4} -\frac{1}{2} \ln 2 ] - [ \frac{\pi}{4} -\frac{1}{2} \ln 2 ]$

$\displaystyle a = \ln 2$

Now, the other part of the question is really easy! Since a parabola is symmetric about its axis, the 2 roots of a quadratic polynomial are symmetric about the point on X-axis corresponding to the maximum/ minimum (minimum in this case; given in the question) i.e. $\ln \sqrt{2}$ is the midpoint of the 2 roots $a$ and $b$ .

So, $\frac{a + b}{2} = \ln \sqrt{2}$

$\frac{ (\ln 2) + b}{2} = \frac{1}{2} (\ln 2)$

$\Rightarrow b = 0$

So, the other root of the quadratic polynomial is $\boxed{0}$ .