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My original Question :

$\displaystyle\int \frac{x^3+1}{(x^3-1) \sqrt {x^4+1}} dx $

What to substitute? How to do it?

I want to add that I have tried doing this using substitutions as :

$x = \frac{1}{t}$ and considered the equation $\displaystyle \frac{1-tan^2 \theta}{1+tan^2 \theta} = \cos 2\theta$ but none of these came to any use.

Also I'm aware of a brother of this problem : $\displaystyle\int \frac{x^2+1}{(x^2-1) \sqrt {x^4+1}} dx$

which also seems like one of Elliptic family as found here : How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$

So now I want to re-phrase my question into some general one :

What is the general solution of $\displaystyle\int \frac{x^n+1}{(x^n-1) \sqrt {x^4+1}} dx$

(Specially when n is odd.)

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    $\begingroup$ I am afraid that, whatever you do, you will face soe nasty elliptic integrals. $\endgroup$ – Claude Leibovici Jul 30 '16 at 3:47
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    $\begingroup$ What's reason of downvoting ? $\endgroup$ – null Jul 30 '16 at 3:48
  • $\begingroup$ @ClaudeLeibovici and I don't know Eliptic integrals.. $\endgroup$ – null Jul 30 '16 at 3:53
  • $\begingroup$ Hint: Change variable first to $y = x+1/x$ and then to $z$ where $y = \frac{z+1/z}{\sqrt{2}}$. $\endgroup$ – achille hui Jul 30 '16 at 4:23
  • $\begingroup$ @achillehui I'm not sure how to implement this $y$ for n=odd cases,(which is my original q) in even cases this comes beautiful. Can you plz elaborate into an answer ? $\endgroup$ – null Jul 30 '16 at 4:36
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First, introduce variables $y, z$ such that $x + x^{-1} = y = \frac{1}{\sqrt{2}}(z+z^{-1})$, we have

$$\begin{align} & \frac{x+1}{x-1}\frac{dx}{\sqrt{x^4+1}} = \frac{x+1}{x-1}\frac{1}{\sqrt{x^2+x^{-2}}}\frac{dx}{x} = \frac{x+1}{x-1}\frac{1}{\sqrt{(x+x^{-1})^2-2}}\frac{d(x+x^{-1})}{x-x^{-1}}\\ = & \frac{dy}{(y-2)\sqrt{y^2-2}}\\ = & \frac{\sqrt{2}}{z+z^{-1}-2\sqrt{2}}\frac{d(z+z^{-1})}{\sqrt{(z+z^{-1})^2-4}} = \frac{\sqrt{2}}{z+z^{-1}-2\sqrt{2}}\frac{d(z+z^{-1})}{z-z^{-1}} = \frac{\sqrt{2}dz}{z^2-2\sqrt{2}z+1} \end{align} $$ Next, for any $n > 1$, we have

$$\frac{x^n - 1}{x-1} = x^{\frac{n-1}{2}}\frac{x^{\frac{n}{2}} - x^{-\frac{n}{2}}}{x^{\frac12}-x^{-\frac12}} = x^{\frac{n-1}{2}} U_{n-1}\left(\frac{x^{\frac12} + x^{-\frac12}}{2}\right)$$ where $U_m(x)$ is the $m^{th}$ Chebyshev polynomial of the $2^{nd}$ kind. When $n = 2k+1$ is an odd number $U_{n-1}(t)$ is an even polynomial of degree $2k$. This means there is a polynomial $f_k(t)$ of degree $k$ such that

$$\frac{x^{2k+1} - 1}{x-1} = x^{k}f_k\left((x^{\frac12} + x^{-\frac12})^2\right) = x^{k} f_k(y+2)$$

Replace $x$ by $-x$, we get $$\frac{x^{2k+1} + 1}{x+1} = (-x)^k f_k(-y+2)$$

Combine these, we find $$\frac{x^{2k+1}+1}{x^{2k+1}-1} = \frac{x+1}{x-1} g_k(y)$$ where $\displaystyle\;g_k(y) = (-1)^k\frac{f_k(2-y)}{f_k(2+y)}$ is a rational function in $y$.

As a result, when $n = 2k+1$ is odd, the integral can be transformed to a integral over a rational function in $z$.

$$\mathcal{I}_n \stackrel{def}{=}\int \frac{x^n+1}{x^n-1}\frac{dx}{x^4+1} = \int g_k\left(\frac{z + z^{-1}}{\sqrt{2}}\right)\frac{\sqrt{2}{dz}}{z^2-2\sqrt{2}z+1}$$

For example, when $n = 3$, $U_{2k}(t) = 4t^2 - 1 \implies f_k(t) = t - 1$. This implies

$$g_1(y) = (-1)^1\frac{(2-y)-1}{(2+y)-1} = \frac{y-1}{y+1}$$

and the integral becomes

$$\begin{align}\mathcal{I}_3 &=\int \frac{y-1}{(y+1)(y-2)}\frac{dy}{\sqrt{y^2-2}} = \frac13 \int \left(\frac{2}{y+1} + \frac{1}{y-2}\right)\frac{dy}{\sqrt{y^2-2}}\\ &= \frac{\sqrt{2}}{3}\int \left(\frac{2}{z^2 + \sqrt{2}z + 1} + \frac{1}{z^2 - 2\sqrt{2}z + 1}\right) dz\\ &= \frac{4}{3}\tan^{-1}(1+\sqrt{2}z) + \frac{1}{3\sqrt{2}}\log\left(\frac{z-\sqrt{2}-1}{z-\sqrt{2}+1}\right) + \text{const...} \end{align} $$

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  • $\begingroup$ Thank you for this incredible answer. I wonder how did this substitution and formulations strike you !!! $\endgroup$ – null Jul 30 '16 at 6:37
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    $\begingroup$ @noob if you see an integral over an algebraic function, one thing one can do is rewrite it to the form $\int f(x) \frac{dx}{x}$ and check whether $f(x)$ has any symmetry under $x \to \text{const}_1 x^{-1}$. If yes, then it is possible to simplify the integral by a change of variable to $x \pm \text{const}_2 x^{-1}$. $\endgroup$ – achille hui Jul 30 '16 at 6:47
  • $\begingroup$ Are these things taught in college? Like these Chebyshev and all ? $\endgroup$ – null Jul 30 '16 at 6:53
  • $\begingroup$ Can u suggest me some book to learn good ways of integrations (difficult ones)? $\endgroup$ – null Jul 30 '16 at 6:55

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