Answer
See explanation.
Work Step by Step
(a) The function $f(x)=\frac{1}{x^{2}+1}$ is always positive, which means integration is always positive also$I>0$
And then
(b) $\frac{1}{u}=x$ and $-\frac{1}{u^{2}} d u=d x$
Replacing $u$ with $x$ then gives
$I=-\int_{-1}^{1} \frac{1}{u^{2}}\cdot \frac{1}{\frac{1}{u^{2}}+1} d u=-\int_{-1}^{1} \frac{1}{u^{2}+1} d u$
This replacement should not be used because $ u $ does not exist if $x= 0$, which is on the interval [-1,1]
$I=-\int_{-1}^{1} \frac{1}{1+x^{2}} d x=-I$
