Integral with a fixed parameter For a parameter $a\ge 1$
How to calculate 

$$\int_0^1 \frac{2\sqrt2}{(\sqrt2+1)(1-u^2)+2au}\ du$$?

I note $p(u) = (\sqrt{2}+1)(1-u^2)+2au$
The roots of $p$ are 
$$u=\frac{-2a-2\sqrt{a^2+(\sqrt2+1)^2}}{-2(\sqrt2+1)}=\frac{-a-\sqrt{a^2+(\sqrt2+1)^2}}{-(\sqrt2+1)}=\frac{-a-\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}$$
and
$$u=\frac{-2a+2\sqrt{a^2+(\sqrt2+1)^2}}{-2(\sqrt2+1)}=\frac{-a+\sqrt{a^2+(\sqrt2+1)^2}}{-(\sqrt2+1)}=\frac{-a+\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}$$ 
So 
$$\begin{align*}\frac{1}{(\sqrt2+1)(1-u^2)+2au}
&=\frac{1}{-(\sqrt2+1)\left(u-\frac{-a-\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}\right)\left(u-\frac{-a+\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}\right)}\\
&=\frac{1-\sqrt{2}}{\left(u-\frac{-a-\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}\right)\left(u-\frac{-a+\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}\right)}\\
&=\frac{1-\sqrt{2}}{\left(u+\frac{-a-\sqrt{a^2+2\sqrt2+3}}{(\sqrt2+1)}\right)\left(u+\frac{-a+\sqrt{a^2+2\sqrt2+3}}{(\sqrt2+1)}\right)}\\
&=\frac{2a}{u+\frac{-a-\sqrt{a^2+2\sqrt2+3}}{(\sqrt2+1)}}+\frac{2a}{u+\frac{-a+\sqrt{a^2+2\sqrt2+3}}{(\sqrt2+1)}}\end{align*}$$
What I wrote is it right? and how can I continue? 
Thanks 
 A: Some Preliminary Steps:
$\require{amsmath}$
$\DeclareMathOperator\arctanh{arctanh}$
$$\begin{aligned}
&(\sqrt2+1)(1-u^2)+2au=-(\sqrt2+1)\left(u^2-\frac{2au}{\sqrt2+1}-1\right)=\\&-(\sqrt2+1)\left(u^2-\frac{2au}{\sqrt2+1}+\frac{a^2}{(\sqrt2+1)^2}-\left(1+\frac{a^2}{(\sqrt2+1)^2}\right)\right)=\\&-(\sqrt2+1)\left(\left(u-\frac{a}{\sqrt2+1}\right)^2-\left(1+\frac{a^2}{(\sqrt2+1)^2}\right)\right)=\\&-(\sqrt2+1)\left(1+\frac{a^2}{(\sqrt2+1)^2}\right)\left(\left(\frac{(\sqrt{2}+1)u-a}{\sqrt{a^2+2\sqrt2+3}}\right)^2-1\right).
\end{aligned}$$

Your integral can now be expressed as:
$$-\frac{2\sqrt2(\sqrt2+1)}{a^2+2\sqrt2+3}\operatorname{\LARGE\int}\frac{1}{\left(\frac{(\sqrt{2}+1)u-a}{\sqrt{a^2+2\sqrt2+3}}\right)^2-1}\,du.$$
Let
$$\begin{aligned}
& s =\frac{(\sqrt{2}+1)u-a}{\sqrt{a^2+2\sqrt2+3}}\\
& ds=\frac{\sqrt{2}+1}{\sqrt{a^2+2 \sqrt{2}+3}}\,du
\end{aligned}$$
and by substituting for $s$, the integral becomes
$$\frac{2\sqrt2}{\sqrt{a^2+2\sqrt2+3}}\operatorname{\LARGE\int}\frac{1}{1-s^2}\,ds=\frac{2\sqrt2\,\arctanh(s)}{\sqrt{a^2+2\sqrt2+3}}+C=\frac{2\sqrt2\,\arctanh\left(\frac{(\sqrt{2}+1)u-a}{\sqrt{a^2+2\sqrt2+3}}\right)}{\sqrt{a^2+2\sqrt2+3}}+C$$
Now, to find the indefinite just plug in the endpoints, to obtain:
$$\boxed{\frac{2 \sqrt{2} \left(\arctanh\left(\frac{\sqrt{2}+1-a}{\sqrt{a^2+2
   \sqrt{2}+3}}\right)+\arctanh\left(\frac{a}{\sqrt{a^2+2
   \sqrt{2}+3}}\right)\right)}{\sqrt{a^2+2 \sqrt{2}+3}}}$$
