Is there an elegant way to evaluate $ I={ \int \sqrt[8]{\frac{x+1}{x}} \ \mathrm{d}x}$? Is there an elegant way to evaluate the following integral?  

$$ I={ \int \sqrt[8]{\dfrac{x+1}{x}} \ \mathrm{d}x}$$   

This seems to me a very lengthy question, yet it was given in my weekly worksheet, so there must be an elegant solution.  
Any help will be appreciated.   

Just to rework @H. R.'s solution by hand. First off setting
$$t=\frac{x+1}x\ge0$$
was definitely a good step. Solving for $x$,
$$x=\frac1{t-1}$$
Note that this implies that either $0\le t<1$ or $t>1$, a property which we will use later to simplify the result. Then
$$\int\sqrt[8]{\frac{x+1}x}dx=-\int\frac{t^{1/8}}{(t-1)^2}dt=-8\int\frac{u^8}{(u^8-1)^2}du$$
Where we have let $t=u^8$, just as @H. R. Further, observe that
$$\frac d{du}\left(\frac u{u^8-1}\right)=\frac{-7u^8-1}{(u^8-1)^2}$$
So that
$$\int\sqrt[8]{\frac{x+1}x}dx=\int\left[\frac d{du}\left(\frac u{u^8-1}\right)-\frac1{u^8-1}\right]du=\frac u{u^8-1}-\int\frac{du}{u^8-1}=\frac u{u^8-1}-J$$
Then partial fractions is easy:
$$\frac1{u^8-1}=\frac1{\prod_{k=0}^7(u-\omega_k)}=\sum_{k=0}^7\frac{A_k}{u-\omega_k}$$
Where $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$ and $\theta_k=\frac{\pi k}4$. Evaluating
$$\lim_{u\rightarrow\omega_k}\frac{u-\omega_k}{u^8-1}=\lim_{u\rightarrow\omega_k}\frac1{8u^7}=\frac1{8\omega_k^7}=\frac{\omega_k}8=\lim_{u\rightarrow\omega_k}\sum_{j=0}^7A_j\frac{u-\omega_k}{u-\omega_j}=\sum_{j=0}^7A_j\delta_{jk}=A_k$$
So
$$\begin{align}J&=\frac18\sum_{k=0}^7\int\frac{\omega_k}{u-\omega_k}du=\frac18\int\left[\frac1{u-1}-\frac1{u+1}+\sum_{k=1}^3\frac{2u\cos\theta_k-2}{u^2-2u\cos\theta_k+1}\right]du\\
&=\frac18\int\left[\frac1{u-1}-\frac1{u+1}+2\sum_{k=1}^3\frac{(u-\cos\theta_k)\cos\theta_k-\sin^2\theta_k}{(u-\cos\theta_k)^2+\sin^2\theta_k}\right]du\\
&=\frac18\left[\ln\left|\frac{u-1}{u+1}\right|+\sum_{k=1}^3\left(\cos\theta_k\ln\left(u^2-2u\cos\theta_k+1\right)-2\sin\theta_k\tan^{-1}\left(\frac{u-\cos\theta_k}{\sin\theta_k}\right)\right)\right]+C\end{align}$$
We may apply the values $\cos\theta_1=-\cos\theta_3=\sin\theta_1=\sin\theta_3=\frac1{\sqrt2}$, $\cos\theta_2=0$, and $\sin\theta_2=0$ to arrive at
$$\begin{align}J&=\frac18\left[\ln\left|\frac{u-1}{u+1}\right|+\frac1{\sqrt2}\ln\left(\frac{u^2-\sqrt2u+1}{u^2+\sqrt2u+1}\right)\right.\\
&\left.-\sqrt2\tan^{-1}\left(\sqrt2u-1\right)-\sqrt2\tan^{-1}\left(\sqrt2u+1\right)-2\tan^{-1}u\right]+C\end{align}$$
Now we can add two of those arctangents because $u$ can't take on values that make the new denominator $0$ nor change the sign of the numerator when the denominator is negative:
$$\begin{align}J&=\frac18\left[\ln\left|\frac{u-1}{u+1}\right|+\frac1{\sqrt2}\ln\left(\frac{u^2-\sqrt2u+1}{u^2+\sqrt2u+1}\right)-\sqrt2\tan^{-1}\left(\frac{\sqrt2u}{1-u^2}\right)-2\tan^{-1}u\right]+C\end{align}$$
We now have the solution   

$$\begin{align}\int\sqrt[8]{\frac{x+1}x}dx&=\frac{\sqrt[8]{\frac{x+1}x}}{\frac{x+1}x-1}-\frac18\left[\ln\left|\frac{\sqrt[8]{\frac{x+1}x}-1}{\sqrt[8]{\frac{x+1}x}+1}\right|+\frac1{\sqrt2}\ln\left(\frac{\sqrt[4]{\frac{x+1}x}-\sqrt2\sqrt[8]{\frac{x+1}x}+1}{\sqrt[4]{\frac{x+1}x}+\sqrt2\sqrt[8]{\frac{x+1}x}+1}\right)\right.\\
&\left.-\sqrt2\tan^{-1}\left(\frac{\sqrt2\sqrt[8]{\frac{x+1}x}}{1-\sqrt[4]{\frac{x+1}x}}\right)-2\tan^{-1}\sqrt[8]{\frac{x+1}x}\right]+C\end{align}$$  

Which matches the Mathematica solution.
