Answer
$a.\quad 0$
$ b.\quad \displaystyle \frac{1}{8}$
Work Step by Step
$a.$
Applying th.7, let $u=g(r)=4+r^{2}$ (a continuous function).
Then, $\displaystyle \quad du=g'(r)dt=2rdr,\qquad rdr=\frac{du}{2}$
and, the borders change to $g(-1)=5$ and $g(1)=5$.
$\displaystyle \int_{0}^{1}\frac{5r}{(4+r^{2})^{2}}dr=\int_{5}^{5}\frac{5}{u^{2}}\cdot\frac{du}{2}$
$=0$
(zero width integral)
$b.$
Using the same substitution, $u=g(r)=4+r^{2}, \ \ du=2rdr$
here, the borders change to $g(0)=4$ and $g(1)=5$.
$\displaystyle \int_{0}^{1}\frac{5r}{(4+r^{2})^{2}}dr=\int_{4}^{5}\frac{5}{u^{2}}\cdot\frac{du}{2}=\frac{5}{4}\int_{4}^{5}u^{-2}du$
$=\displaystyle \frac{5}{4}\left[ \frac{u^{-2+1}}{-2+1} \right]_{4}^{5}$
$=\displaystyle \frac{5}{4}\left[ -u^{-1} \right]_{4}^{5}$
$=\displaystyle \frac{5}{4}(-\frac{1}{5}+\frac{1}{4})$
$=\displaystyle \frac{5}{4}\cdot\frac{1}{20}$
$=\displaystyle \frac{1}{8}$
