How to integrate $\int \frac{1+x^{11}}{1-x^{11}}dx$?

Question

How to integrate $$\int \frac{1+x^{11}}{1-x^{11}}dx$$ ?

I am trying to solve this integral for one week but can't.

Let me tell how I started

$$\int\frac{1+x^{11}}{1-x^{11}}dx=\int\frac{1}{1-x^{11}}dx+\int\frac{x^{11}}{1-x^{11}}dx$$

Now I know that

$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$$ when $|x|<1$

Therefore we can also say that

$$\frac{1}{1-x^{11}}=\sum_{n=0}^{\infty}x^{11n}$$ when $|x^{11}|<1$

Therefore, the integral will become

$$\int \frac{1+x^{11}}{1-x^{11}}dx=\int\sum_{n=0}^{\infty}x^{11n}dx$$+$$ \int x^{11}\sum_{n=0}^{\infty}x^{11n}dx$$

But after this step I can't understand how to integrate.

Also I don't know whether I can apply the formula of $\frac{1}{1-x^{11}}$ here because the formula is applicable only for $|x^{11}|<1$.

Hence, kindly clear my doubt.

Answer 1

Let $x= -t$. Then we have

$$\begin{align} \int \frac{1+x^{11}}{1-x^{11}}\,dx&=\int \frac{t^{11}-1}{1+t^{11}}\,dt\\\\ &=1-2\int \frac{1}{1+t^{11}}\,dt\\\\ \end{align}$$

Now, in THIS ANSWER, I showed that the indefinite integral $I_n=\int \frac1{1+x^n}\,dx$, $n\in \mathbb{N_{+}}$ is given by

$$\begin{align} I_n&=\int\frac{1}{x^n+1}dx\\\\ &=-\frac1n\sum_{k=1}^n\left(\frac12 x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)+C \end{align}$$

where $x_{kr}$ and $x_{ki}$ are are given by

$$x_{kr}=\text{Re}\left(x_k\right)=\cos \left(\frac{(2k-1)\pi}{n}\right)$$

$$x_{ki}=\text{Im}\left(x_k\right)=\sin \left(\frac{(2k-1)\pi}{n}\right)$$

Now, set $n=11$. Can you finish?

Answer 2

This is just one method. You could do the long division:

$$\int\left(-1+\frac{2}{1-x^{11}}\right)\,dx=-x+2\int\frac{1}{1-x^{11}}\,dx$$

Now you only have to address one of the two integrals that your outline introduced. The integrand is a rational function with the degree of the numerator less than that of the denominator. So it is not fun, but you can use partial fraction decomposition, working with complex numbers to make the notation easier:

$$\begin{align} \int\frac{1}{x^{11}-1}\,dx &=\int\sum_{k=-5}^{5}\frac{A_k}{x-\omega^k}\,dx \end{align}$$

where $\omega$ is a primitive $11$th root of $1$. You would have to work out what the constants $A_k$ are using linear algebra or something.

Then you have

$$\begin{align} \int\frac{1}{x^{11}-1}\,dx &=\sum_{k=-5}^{5}A_k\ln(x-\omega^k)\,dx\\ &=A_0\ln(x-1)+\sum_{k=1}^5\ln\left( \left( x-\omega^k \right)^{A_k}\left( x-\omega^{-k} \right)^{A_{-k}}\right) \end{align}$$

I'm pretty sure there will be a relationship between $A_k$ and $A_{-k}$ that make those arguments to $\ln$ become real, expressible using trig expressions with $2\pi/11$ in them.

Putting this all together you would have an antiderivative for the original integrand.