The value of $\int\frac{1}{(2ax+x^2)^{3/2}}dx$ is:

$\frac{-1}{a^2}\frac{x+a}{\sqrt{x^2+2ax}}+c$

$\int\frac{1}{(2ax+x^2)^{3/2}}dx$ [Put x + a = a sec θ ⇒ dx = a sec θ.tan θ.dθ]

$\int\frac{a\sec θ\tan θ}{a^3(\tan^2θ)^{3/2}}dθ=\frac{1}{a^2}\int\frac{1}{\cos θ}.\frac{\cos^2θ}{\sin^2θ}dθ=\frac{1}{a^2}\int\frac{\cos θ}{\sin^2 θ}dθ=\frac{1}{a^2}.\frac{-1}{\sin θ}=-\frac{1}{a^2}.\frac{x+a}{\sqrt{x^2+2ax}}+C$

(Make a triangle for $\cos θ=\frac{a}{x+a}$ and find $\sin θ$ as $\frac{x+a}{\sqrt{x^2+2ax}}$)