Integrate the following : $\displaystyle\int \frac{x^n+1}{(x^n-1) \sqrt {x^4+1}} dx$ ; n is odd.

Question

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.)

Answer 1

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

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

Answer 2

Here's an alternative answer to the "original question":

$$\begin{align*} &\int \frac{x^3+1}{(x^3-1) \sqrt {x^4+1}} \, dx \\[2ex] &= \int -\frac1y\cdot\frac{\left(\frac{1-y}{1+y}\right)^2-\frac{1-y}{1+y}+1}{\left(\frac{1-y}{1+y}\right)^2+\frac{1-y}{1+y}+1} \cdot \frac{1}{\sqrt{\left(\frac{1-y}{1+y}\right)^4+1}} \cdot \frac{-2\,dy}{(1+y)^2} \tag1 \\[1ex] &= 2 \int \frac{3y^2+1}{y\left(y^2+3\right)}\cdot\frac{dy}{\sqrt{2y^4+12y^2+2}} \\[1ex] &= \frac{\sqrt2}3 \int \frac1y \left(1 + \frac{8y^2}{y^2+3}\right) \frac{dy}{\sqrt{\left(y^2+3\right)^2-8}} \\[2ex] &= \frac1{\sqrt2} \int \left(3-\frac8z\right) \frac{dz}{(z-3)\sqrt{z^2-8}} \tag2 \\[1ex] &= \frac1{3\sqrt2} \int \left(\frac1{z-3}+\frac8z\right) \frac{dz}{\sqrt{z^2-8}} \\[2ex] &= \frac1{3\sqrt2} \int \left(\frac1{-\frac{w^2+8}{2w}-3}+\frac8{-\frac{w^2+8}{2w}}\right) \frac{\frac4{w^2}-\frac12}{w-\frac{w^2+8}{2w}} \, dw \tag3 \\[1ex] &= \frac{1}{3\sqrt2} \int \left(\frac1{w+2} - \frac1{w+4} + \frac {16}{w^2+8}\right) \, dw \end{align*}$$

---

• $(1)$ : substitute $y=\dfrac{1-x}{1+x} \iff x=\dfrac{1-y}{1+y}$; simplify and expand into partial fractions along the way
• $(2)$ : substitute $y=\sqrt{z-3} \iff z=y^2+3$
• $(3)$ : substitute $w=\sqrt{z^2-8}-z \iff z=-\dfrac{w^2+8}{2w}$