Answer
$$\int^{1}_{-1}\frac{5r}{(4+r^2)^2}dr=0$$ $$\int^{1}_{0}\frac{5r}{(4+r^2)^2}dr=\frac{1}{8}$$
Work Step by Step
Part a) $$A=\int^{1}_{-1}\frac{5r}{(4+r^2)^2}dr$$
Set $u=4+r^2$, we then have $$du=2rdr$$ $$rdr=\frac{1}{2}du$$
For $r=-1$, we have $u=5$ and for $r=1$, we have $u=5$.
Therefore, $$A=\frac{1}{2}\int^{5}_5\frac{5}{u^2}du=0$$
Part b) $$A=\int^{1}_{0}\frac{5r}{(4+r^2)^2}dr$$
We carry out a similar substitution as in part a).
For $r=0$, we have $u=4$ and for $r=1$, we have $u=5$.
Therefore, $$A=\frac{1}{2}\int^{5}_4\frac{5}{u^2}du$$ $$A=\frac{5}{2}\Big(-\frac{1}{u}\Big)\Big]^5_4$$ $$A=-\frac{5}{2}\Big(\frac{1}{5}-\frac{1}{4}\Big)$$ $$A=-\frac{5}{2}\Big(-\frac{1}{20}\Big)=\frac{1}{8}$$
