I'm trying to integrate
$$I=\int_{1}^{\infty}\frac{(x-1)}{(x+1)\sqrt{(a+x)}}dx$$
Where $a$ is a constant.
$\textbf{My Attempt:}$
Letting $u=\sqrt{a+v}$ and ignoring the limits for now gives
$$\begin{align}I&=2\int\bigg(\frac{x-1}{x+1}\bigg)du \\ &=2\int\bigg(\frac{a-u^2+1}{a-u^2-1}\bigg)du \\ &=2\int\bigg(1-\frac{2}{-a+u^2+1}\bigg)du \\ &=-\frac{4}{1-a}\int\frac{1}{\bigg(\frac{u^2}{1-a}+1\bigg)}du+2\int du \end{align}$$ Now let $p=\frac{u}{\sqrt{1-a}}$ to obtain
$$\begin{align}I&=-\frac{4}{\sqrt{1-a}}\int\frac{dp}{p^2+1}+2\int du \\ &= 2u -\frac{4}{\sqrt{1-a}}\tanh^{-1}{\bigg(\frac{u}{\sqrt{1-a}}\bigg)}. \end{align}$$
That means now that
$$\int_{1}^{\infty}\frac{(x-1)}{(x+1)\sqrt{(a+x)}}dx = \bigg[2u -\frac{4}{\sqrt{1-a}}\tanh^{-1}{\bigg(\frac{u}{\sqrt{1-a}}\bigg)}\bigg]_{\sqrt{a+1}}^{\infty}$$
which doesn't converge. Am I missing something here? I should be able to find a closed form solution. I am thinking a better substitution could be made. Any help much appreciated.