Answer
$a.\quad 4$
$ b.\quad 0$
Work Step by Step
$a.$
Applying th.7, let $u=g(x)=1+x^{2}$ (a continuous function).
Then, $\quad du=g'(v)dv=2xdx,\qquad 4xdx=2\cdot du$
and, the borders change to $g(0)=1$ and $g(\sqrt{3})=4$.
$\displaystyle \int_{0}^{\sqrt{3}}\frac{4x}{\sqrt{x^{2}+1}}dr=\int_{1}^{4}\frac{2du}{u^{1/2}}=2\int_{1}^{4}u^{-1/2}du$
$=2\displaystyle \left[ \frac{u^{-1/2+1}}{-1/2+1} \right]_{1}^{4}$
$=2\displaystyle \left[ \frac{u^{1/2}}{1/2} \right]_{1}^{4}$
$=4\left[ u^{1/2} \right]_{1}^{4}$
$=4(2-1)$
$=4$
$b.$
Using the same substitution, $u=1+x^{2},\ du=2xdx$,
the borders here change to
$g(-\sqrt{3})=4$ and $g(\sqrt{3})=4$.
$\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}\frac{4x}{\sqrt{x^{2}+1}}dr=\int_{4}^{4}\frac{2du}{u^{1/2}} = 0,$
(zero width integral)
