Substitution: why can we ignore the absolute value in this case? I am trying to solve this integral:
$$\int\frac{dx}{x(3+x^2)\sqrt{1-x^2}}$$
We can use substitution:$$1-x^2=u^2$$ and $$-x dx=u du$$
Which gives us:
$$-\int\frac{udu}{(1-u^2)(4-u^2)|u|}$$
Now my calculus-book then just proceeds saying this integral equals:
$$-\int\frac{du}{(1-u^2)(4-u^2)}$$
Question: why can we just ignore the absolute value in this case? 
Couldn't $u$ be both $+\sqrt{1-x^2}$ or $-\sqrt{1-x^2}$? I was expecting to make two separate cases, one for $u<0$ and one for $u\ge0$. Why is this not so?

EDIT: Perhaps the mistake that I made is that the real substitution isn't $1-x^2=u^2$, but $u=\sqrt{1-x^2}$, which is positive?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\int{\dd x \over x\pars{3 + x^{2}}\root{1 -x^{2}}} =
{1 \over 3}\int{\dd x \over x\root{1 - x^{2}}} -
{1 \over 3}\int{x \over \pars{3 + x^{2}}\root{1 - x^{2}}}\,\dd x:\ {\large ?}}$

\begin{align}
{1 \over 3}\int{\dd x \over x\root{1 - x^{2}}} &
\,\,\,\stackrel{x\ =\ 1/t}{=}\,\,\,
-\,{1 \over 3}\int{\dd t \over \root{t^{2} - 1}}
\,\,\,\stackrel{t\ =\ \sec\pars{\theta}}{=}\,\,\,
-\,{1 \over 3}\int\sec\pars{\theta}\,\dd \theta
\\[5mm] & =
-\,{1 \over 3}\,\ln\pars{\verts{\sec\pars{\theta} + \tan\pars{\theta}}} =
-\,{1 \over 3}\,\ln\pars{\verts{t + \root{t^{2} - 1}}}
\\[5mm] & =
{1 \over 3}\ln\pars{\verts{x \over 1 + \root{1 - x^{2}}}}
\\[5mm] & =
{1 \over 6}\bracks{\ln\pars{\verts{x \over 1 + \root{1 - x^{2}}}} +
\ln\pars{\verts{1 - \root{1 - x^{2}} \over x}}}
\\[5mm] & =
{1 \over 6}\,\ln\pars{\verts{1 - \root{1 - x^{2}} \over
1 + \root{1 - x^{2}}}}
\end{align}

\begin{align}
{1 \over 3}\int{x \over \pars{3 + x^{2}}\root{1 - x^{2}}}\,\dd x &
\,\,\,\stackrel{x\ =\ \root{1 - t^{2}}}{=}\,\,\,
{1 \over 3}\int{\dd t \over t^{2} - 4} =
{1 \over 12}\int\pars{{1 \over t - 2} - {1 \over t + 2}}\,\dd t
\\[5mm] & =
{1 \over 12}\ln\pars{\verts{t - 2 \over t + 2}} =
\bbx{-\,{1 \over 12}\ln\pars{\verts{\root{1 - x^{2}} + 2 \over
\root{1 - x^{2}} - 2}}}
\end{align}

\begin{align}
&\int{\dd x \over x\pars{3 + x^{2}}\root{1 -x^{2}}}
\\[5mm] = &\
\bbx{%
{1 \over 6}\ln\pars{\verts{1 - \root{1 - x^{2}} \over 1 + \root{1 - x^{2}}}} +
{1 \over 12}\ln\pars{\verts{\root{1 - x^{2}} + 2 \over \root{1 - x^{2}} - 2}} + \mbox{a constant}}
\end{align}
