Substitution: why can we ignore the absolute value in this case?

Question

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?

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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?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int{\dd x \over x\pars{3 + x^{2}}\root{1 -x^{2}}} = {1 \over 3}\int{\dd x \over x\root{1 - x^{2}}} - {1 \over 3}\int{x \over \pars{3 + x^{2}}\root{1 - x^{2}}}\,\dd x:\ {\large ?}}$

\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}

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\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}

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\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}