Answer
$$\int^{\ln5}_{0}e^r(3e^r+1)^{-3/2}dr=\frac{1}{6}$$
Work Step by Step
$$A=\int^{\ln5}_{0}e^r(3e^r+1)^{-3/2}dr$$
We set $u=3e^r+1$, which means $$du=3e^rdr$$ $$e^rdr=\frac{1}{3}du$$
For $r=\ln5$, $$u=3e^{\ln5}+1=3\times5+1=16$$
For $r=0$, $$u=3e^{0}+1=3\times1+1=4$$
Therefore, $$A=\frac{1}{3}\int^{16}_{4}u^{-3/2}du$$ $$A=\frac{1}{3}\times\frac{u^{-1/2}}{-\frac{1}{2}}\Big]^{16}_{4}$$ $$A=-\frac{2}{3\sqrt u}\Big]^{16}_{4}$$ $$A=-\Big(\frac{2}{3\sqrt{16}}-\frac{2}{3\sqrt4}\Big)$$ $$A=-\Big(\frac{1}{6}-\frac{1}{3}\Big)$$ $$A=\frac{1}{6}$$
