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The integral can be represented as $$ \int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx= \int \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx $$

Substitution $$t=\frac{1+x}{1-x}\Rightarrow x=\frac{t-1}{t+1}\Rightarrow dx=\frac{2}{(t+1)^2}dt\Rightarrow \int\limits \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx=2\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$$

What substitution to use for solving the integral $\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$?

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    $\begingroup$ Why not using $x=\cos t$ instead? $\endgroup$ – Pierpaolo Vivo Feb 9 '16 at 10:49
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No substitutions: $$ \int\left(\frac{1}{\sqrt{1-x^2}}-\frac{-x}{\sqrt{1-x^2}}\right)\,dx =\arcsin x-\sqrt{1-x^2}+c $$

You can also do that way; continue with $u=\sqrt{t}$, so $t=u^2$ and $dt=2u\,du$; so you get $$ \int\frac{4u^2}{(u^2+1)^2}\,du= \int 2u\cdot\frac{2u}{(u^2+1)^2}\,du $$ Noticing that $2u$ is the derivative of $u^2+1$ you can use integration by parts $$ =2u\cdot\left(-\frac{1}{u^2+1}\right)- \int2\left(-\frac{1}{u^2+1}\right)\,du =2\arctan u-\frac{2u}{u^2+1} $$ Do the back substitutions.

Alternative method: set $x=\cos4t$, so you have $$ \sqrt{\frac{1+\cos4t}{1-\cos4t}}=\frac{\cos2t}{\sin2t} $$ and the integral becomes $$ -8\int\cos^22t\,dt=-4\int(1+\cos t)\,dt $$

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  • $\begingroup$ "No substitutions": did you just guess that $\int \frac{-x}{1-x^2} dx = \sqrt{1-x^2}$, or did you do the substitution (even just in your head) $y = 1-x^2$...? $\endgroup$ – Najib Idrissi Feb 9 '16 at 10:57
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    $\begingroup$ @NajibIdrissi The derivative of $1-x^2$ is $-2x$. $\endgroup$ – egreg Feb 9 '16 at 11:00
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    $\begingroup$ Really?! Thanks for the information! ... More seriously, I think most students will just slap a $\pm 2^k$ coefficient (for some $k \in \mathbb{Z}$) and not bother taking the derivative to verify their solution. I'm not sure that "no substitutions" is good advice. $\endgroup$ – Najib Idrissi Feb 9 '16 at 11:02
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    $\begingroup$ @NajibIdrissi Decomposing the fraction should always a first attempt before undertaking substitutions; the presence of $x^2$ and $x\,dx$ should make a bell ring. $\endgroup$ – egreg Feb 9 '16 at 11:04
  • $\begingroup$ Of course it should make a bell ring, but there's a big step between being able to guess the solution without making a mistake and having to do a substitution to find the correct solution. Anyway, I don't care enough about this to continue this discussion. $\endgroup$ – Najib Idrissi Feb 9 '16 at 11:06
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We can also use the substitution $u=\sqrt{1-x}$, then $\mathrm{d}u=-\frac{\mathrm{d}x}{2\sqrt{1-x}}$ and $\sqrt{1+x}=\sqrt{2-u^2}$. We will also use $u=\sqrt2\sin(\theta)$ $$ \begin{align} \int\frac{\sqrt{1+x}}{\sqrt{1-x}}\,\mathrm{d}x &=-2\int\sqrt{2-u^2}\,\mathrm{d}u\\ &=-2\int\sqrt2\cos(\theta)\cdot\sqrt2\cos(\theta)\,\mathrm{d}\theta\\ &=-2\int(1+\cos(2\theta))\,\mathrm{d}\theta\\[3pt] &=-2\theta-\sin(2\theta)+C\\[3pt] &=-2\arcsin\left(\sqrt{\frac{1-x}{2}}\right)-\sqrt{1-x^2}+C\\ &=-\arccos(x)-\sqrt{1-x^2}+C\\[6pt] &=\arcsin(x)-\sqrt{1-x^2}+\left(C-\tfrac\pi2\right) \end{align} $$


Explanation of $\boldsymbol{2\arcsin\left(\sqrt{\frac{1-x}{2}}\right)=\arccos(x)}$

Let $\alpha=\arcsin\left(\sqrt{\frac{1-x}2}\right)$, then $x=1-2\sin^2(\alpha)=\cos(2\alpha)$. Thus, $\alpha=\frac12\arccos(x)$. Therefore, $$ 2\arcsin\left(\sqrt{\frac{1-x}2}\right)=\arccos(x) $$

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  • $\begingroup$ I now see that I answered Takahiro Waki's modification of the question, which was subsequently rolled back to the original question. The answer is still correct, but now I see why egreg's answer took the path it did. $\endgroup$ – robjohn Feb 9 '16 at 17:19

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