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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?

GambitSquared
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    That is because $\sqrt A$ denotes the non-negative square root of the non-negative number $A$. – Bernard May 04 '18 at 21:30
  • It is assumed that $u\ge0$. Otherwise it is not well-defined by means of equation $1-x^2=u^2$. – user May 04 '18 at 21:36
  • If all you know is that $1-x^2=u^2$ couldn't $u$ then be both $+\sqrt{1-x^2}$ or $-\sqrt{1-x^2}$? – GambitSquared May 04 '18 at 21:47
  • See the comment above. Of course you are free to choose either of the forms. But you cannot choose both of them. – user May 04 '18 at 21:51
  • @user That was what I was commenting about, I don't see what you mean by not "well-defined" – GambitSquared May 04 '18 at 21:55
  • @Bernard if you are solving $1-x^2=u^2$ for $u$ then there are two solutions right? Why don't we see both solutions back in the substitution? – GambitSquared May 04 '18 at 21:56
  • I mean that the map $x\mapsto u$ should be one-valued. Imagine you are going to compute definite integral from $x_1$ to $x_2$. Could you choose $u_1=\sqrt{1-x_1^2}$, $u_2=-\sqrt{1-x_2^2}$? – user May 04 '18 at 21:58
  • @GambitSquared Bernard hit the nail on the head. Suppose that you were asked to evaluate $\int_{-1}^1 \sqrt{x^2} dx.$ You would assume that if $f(x) = \sqrt{x^2},$ then $f(x)$ is a non-negative function throughout the interval $[-1,1].$ – user2661923 May 04 '18 at 21:59
  • @user2661923 Exactly for that reason you cannot say that $f(x)=\sqrt{x^2}=x$ for the whole domain. For $x<0$ then $f(x)=-x$ – GambitSquared May 04 '18 at 22:05
  • @GambitSquared Agreed. Therefore, by convention $f(x)$ (in my example) is construed to represent $|x|,$ which is the point that Bernard was making in the first place. – user2661923 May 04 '18 at 22:36
  • My answer is here. – ryang Feb 02 '22 at 10:47

1 Answers1

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

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

\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}

\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}
Felix Marin
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  • Are you sure this is the right answer? It doesn't have any real values because of $\sqrt{1-x^2}-2$ My calculus-book gives the answer: $\frac{1}{6}\log |\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}|+\frac{1}{12}\log |\frac{2+\sqrt{1-x^2}}{2-\sqrt{1-x^2}}|+C$ But anyway my real question was why we can get rid of the absolute sign... – GambitSquared May 05 '18 at 09:43
  • @GambitSquared Done. Thanks. – Felix Marin May 05 '18 at 18:54