$\int \frac{(2x+1)}{{(x^2+4x+1)}^{3/2}}dx.$  is equal to (where  $C$ is the constant of integration)

 $$\begin{gathered} \int \frac{2x+1}{{(x^2+4x+1)}^{3/2}}dx=\int \frac{2x-1}{x^3{(1+\frac{4}{x}+\frac{1}{x^2})}^{3/2}}dx \\ =\int \frac{2x^{-2}+x^{-3}}{{(1+\frac{4}{x}+\frac{1}{x^2})}^{3/2}}dx \end{gathered}$$
 Now put  $\frac{1}{x^2}+\frac{4}{x}+1=t$
  $$\begin{gathered} \Rightarrow (-\frac{2}{x^3}-\frac{4}{x^2})dx=dt \\ \therefore I=-\frac{1}{2}\int \frac{1}{t^{3/2}}dt \\ I=\frac{x}{{(x^2+4x+1)}^{1/2}}+c \end{gathered}$$