If $\alpha >1,$ then $\int \frac{dx}{x^2+2\alpha x+1}=$

$\alpha >1,\int \frac{dx}{x^2+2dx+1}$

$\int \frac{dx}{{(x+\alpha )}^2-{\alpha }^2+1}$

$\int \frac{dx}{{(x+\alpha )}^2-({\alpha }^2-1)} \left[\because \alpha >1\Rightarrow {\alpha }^2>1\Rightarrow ({\alpha }^2-1)>0\right]$

$=\int \frac{dx}{{(x+\alpha )}^2-{\left(\sqrt{\alpha -1}\right)}^2}$

$$\begin{gathered} =\frac{1}{2\sqrt{{\alpha }^2-1}}\log \left|\frac{x+\alpha -\sqrt{{\alpha }^2-1}}{x+\alpha +\sqrt{{\alpha }^2-1}}\right|+C \\ \left(\because \int \frac{{\displaystyle 1}}{{\displaystyle x^2+{\alpha }^2}}=\frac{{\displaystyle 1}}{{\displaystyle 2a}}\log \left|\frac{{\displaystyle x-a}}{{\displaystyle x+a}}\right|\right) \end{gathered}$$