How to integrate $\int \frac{\sqrt[n]{x+\frac{1}{x}}}{1-\frac{1}{x^{2}}}dx$ for $n\geq 2$?

Question

Q) How to integrate $$\int \frac{\sqrt[n]{x+\frac{1}{x}}}{1-\frac{1}{x^{2}}}dx$$ for $n\geq 2$?

Ans) My Approach :

When I was in Class $XIITh$ I learnt how to integrate $$\int \frac{1-\frac{1}{x^{2}}}{x+\frac{1}{x}}dx$$ . To Integrate $$\int \frac{1-\frac{1}{x^{2}}}{x+\frac{1}{x}}dx$$ we substitute $(x+\frac{1}{x})=t$ and then $(1-\frac{1}{x^{2}})dx=dt$ . Therefore we can say that

$$\int \frac{1-\frac{1}{x^{2}}}{x+\frac{1}{x}}dx=\ln|x+\frac{1}{x}|+C$$

Similarly we can also integrate:

$$\int \frac{x+\frac{1}{x}}{1-\frac{1}{x^{2}}}dx$$ .

Let $$I=\int \frac{x+\frac{1}{x}}{1-\frac{1}{x^{2}}}dx$$

$$\implies I=\int \frac{x^{3}+x}{x^{2}-1}dx$$

$$\implies I=\int \frac{x(x^{2}-1)+2x}{x^{2}-1}dx$$

$$\implies I=\int (x+\frac{2x}{x^{2}-1})dx$$

$$\implies I= \frac{x^{2}}{2}+\ln|x^{2}-1|+C$$

My doubt :

I am facing problem in integrating

$$\int \frac{\sqrt{x+\frac{1}{x}}}{1-\frac{1}{x^{2}}}dx$$

and $$\int \frac{\sqrt[3]{x+\frac{1}{x}}}{1-\frac{1}{x^{2}}}dx$$

I have also searched the solutions in Wolfram Alpha but I have not understood the way of solution.

For the integral $$\int\frac{\sqrt{x+\frac{1}{x}}}{1-\frac{1}{x^{2}}}dx$$ , as far as I have understood it is giving answer in terms of hypergeometric functions.

And for the integral $$\int\frac{\sqrt[3]{x+\frac{1}{x}}}{1-\frac{1}{x^{2}}}dx$$ Wolfram Alpha is not giving answer in terms of standard mathematical functions.

Please help me out with this integral.

Answer 1

Let $u(x) = \frac{1}{x}$, then $$\int \frac{(x + u)^{1/n}}{1 - u^2} dx$$

Let $(x + u) = (1 - u^2)$, and then $x = 1 - u - u^2$ so \begin{align} x &= \frac{u \pm \sqrt{ 5u^2}}{2}\\ &= \frac{u + |u|\sqrt{5}}{2} \end{align} but $u=1/x$ hence

$$2x = \frac{1}{x} + \left|\frac{1}{x}\right|\sqrt{5}$$ by composition.

That means, in order for us to try and cancel the numerator and denominator— there is a very specific value for $x$ for that to be possible (it is not divisible for all values of $x$). Plus, I think I messed up the quadratic, but the idea is there.

So you can’t divide these terms via this form of factorization.

1. You can try a polynomial expansion: $$(a + b)^n = \sum_{i=0}^n\binom{n}{i} a^ib^{n-i} $$ then try to isolate $(a + b)$ — which shouldn’t work, because there is no equation for factorizing $n$-degree polynomial expansions. I think.

2. $\tan(a + b)$ has a similar form. You can try and find a substitution, but it’ll be hard: $$\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$$

So, let $\tan(a + b) = \tan\left(F(x,u) + G(x,u)\right)$, but idk what we should define for $F$ and $G$, so that we get a clean matching expression.

Also, you still have to consider the $n$-root part of the equation.

You can try Partial Fraction Decomposition. It might be easier. Consult a professional— that is not something I know how to do very well.

I don’t think there is a solution. Even if there is, I don't think it looks important. You can try approximating the solution with some Optimization expression— I recommend not pursuing this problem if you know there is no solution.