$~\int\frac{1}{x+3}\sqrt{\frac{x+2}{x+4}}\, dx~$with using the constraint $~t=\sqrt{\frac{x+2}{x+4}}~$
Solution 1:
First, express $x$ in terms of $t$. Squaring, we get $$t^2=\frac{x+2}{x+4}=1-\frac{2}{x+4}\\x+4=\frac{2}{1-t^2} \\x=\frac{4t^2-2}{1-t^2}\\ \Rightarrow dx=\frac{4t}{(1-t^2)^2}dt$$ Then substitute in the integral: $$\int\frac{t}{3+\frac{4t^2-2}{1-t^2}}\frac{4t}{(1-t^2)^2}\,dt=\int\frac{4t^2}{(1-t^2)(t^2+1)}\,dt$$ The partial fraction decomposition is $$\begin{align}\frac{4t^2}{(1-t^2)(t^2+1)}&= \frac{2(1+t^2)-2(1-t^2)}{(1-t^2)(t^2+1)}\\ &=\frac{2}{1-t^2}-\frac{2}{t^2+1}\\ &=\frac{1}{1-t}+\frac{1}{t+1}-\frac{2}{t^2+1}\end{align} $$ Can you continue from here?