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.)

  • 1
    $\begingroup$ I am afraid that, whatever you do, you will face soe nasty elliptic integrals. $\endgroup$ – Claude Leibovici Jul 30 '16 at 3:47
  • 1
    $\begingroup$ 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

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

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

  • $\begingroup$ Thank you for this incredible answer. I wonder how did this substitution and formulations strike you !!! $\endgroup$ – null Jul 30 '16 at 6:37
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.