Let $\mathrm{I}=\int \frac{d x}{\left(2 a x+x^2\right)^{3 / 2}}=\int \frac{d x}{\left((x+a)^2-a^2\right)^{3 / 2}}$
Let $t=x+a \Rightarrow d t=d x$
$\Rightarrow \mathrm{I}=\int \frac{d t}{\left(t^2-a^2\right)^{3 / 7}}$
Put $t=a \sec p$
$\begin{aligned} & \Rightarrow \frac{d t}{d p}=a \sec p \tan p \Rightarrow d t=a \sec p \tan p d p \\ & \therefore \mathrm{I}=\int \frac{a \sec p \tan p}{\left(a^2 \sec 2 p-a^2\right)^{3 / 2}} d p \\ & =\frac{1}{a^2} \int \frac{\sec p \tan p}{\left(\tan ^2 p\right)^{3 / 2}} d p=\frac{1}{a^2} \int \frac{\cos p}{\sin ^2 p} d p \\ & =-\frac{1}{a^2}\left(\frac{1}{\sin p}\right)+c=-\frac{1}{a^2}\left(\frac{1}{\sin \left(\sec ^{-1} \frac{t}{a}\right)}\right)+c \\ & =-\frac{1}{a^2}\left(\frac{1}{\left(\frac{(a) \sqrt{\frac{t^2}{a^2}}-1}{t}\right)+c=-\frac{1}{a^3}\left(\frac{t}{\sqrt{\frac{t^2}{a^2}-1}}\right)+c}\right. \\ & =-\frac{1}{a^3}\left(\frac{x+a}{\sqrt{\frac{(\mathrm{x}+\mathrm{a})^2}{a^2}-1}}\right)+c=-\frac{1}{a^2}\left(\frac{x+a}{\sqrt{x^2+2 a x}}\right)+c\end{aligned}$