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In order to calculate the values of $a$ and $b$ such we get the minimum possible for:

$$\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}\,dx$$

I got the help of @TheSimpliFire among others to get the respectively $a$ and $b$ here:

Find $a$ and $b$ for which $\int_{0}^{1}( ax+b+\frac{1}{1+x^{2}} )^{2}\,dx$ takes its minimum possible value.

Then, as we found were the partial derivatives of $a$ and $b$ are zero it is not hard to prove the founded $(a,b)$ satisfies that for:

$$D(a,b) = f_{xx}'(a,b)f_{yy}'(a,b)-f_{xy}(a,b)^2 $$

As for $p=(a,b)$;$$D(p) > 0 \land f_{xx}'(p) > 0 \implies \text{minimum}$$ i s satisfied, then $(a,b)$ are locally minimum. My question is how to verify $(a,b)$ is also the global minimum?? Thanks!!!

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May be, we could just use brute force.

Since $$F=\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}\,dx=\frac{a^2}{3}+a (b+\log (2))+\frac{1}{8} \left(8 b^2+4 \pi b+\pi +2\right)$$ $$\frac{\partial F}{\partial a}=\frac{2 a}{3}+b+\log (2)=0 \implies a=-\frac{3}{2} (b+\log (2))$$ Replace in $F$ to get $$F=\frac{1}{8} \left(2+\pi +6 \pi ^2+48 \log ^2(2)-36 \pi \log (2)\right)+ (2 \pi -6 \log (2))b+b^2$$ which is just a quadratic in $b$ and shows a single minimum for $$b=3\log(2)-\pi \implies a=\frac{3}{2} (\pi -4 \log (2))$$ $$\implies F_{min}=\frac{1}{8} \left(2+\pi -2 \pi ^2-24 \log ^2(2)+12 \pi \log (2)\right)$$

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Call your integral $f(a,b)$. You have found a single point ${\bf p}\in{\mathbb R}^2$ with $\nabla f({\bf p})={\bf 0}$. Let $f({\bf p})=:c>0$.

Inspecting the expression defining $f(a,b)$ it is obvious that there is an $M>0$ such that $f(a,b)\geq 2c$ when $a^2+b^2\geq M^2$. The disc $B_M: \>a^2+b^2\leq M^2$ is compact, and $f$ is continuous on $B_M$. Therefore $f$ assumes a minimum $\leq c$ on $B_M$, and this cannot be at a point ${\bf q}\in\partial B_M$. This minimum is therefore a local minimum of $f$ at an interior point of $B_M$, hence $\nabla f({\bf q})={\bf 0}$ at this point. There is only one such point, namey the point ${\bf p}$ found in your calculations.

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