Evaluating :$\int \frac{1}{x^{10} + x}dx$ $$\int \frac{1}{x^{10} + x}dx$$
My solution :
$$\begin{align*}
\int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\
&=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\
&=\ln|x|-\frac{1}{9}\ln|x^9+1|+C
\end{align*}$$
Is there completely different way to solve it ?

Let we generalise the problem with a slightly different way. Consider
$$\int\frac{\mathrm dx}{x^n+x}$$
Making substitution $x=\frac{1}{y}$ and $z=y^{n-1}$ then reverse it back we get
\begin{align}
\int\frac{\mathrm dx}{x^n+x}&=-\int\frac{y^{n-2}}{1+y^{n-1}}\mathrm dy\\[9pt]
&=-\frac{1}{n-1}\int\frac{\mathrm dz}{1+z}\\[9pt]
&=-\frac{\ln |1+z|}{n-1}+C\\[9pt]
&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\ln |x|-\frac{\ln \left|1+x^{n-1}\right|}{n-1}+C}}
\end{align}
In your case
$$\int\frac{\mathrm dx}{x^{10}+x}=\ln |x|-\frac{\ln \left|1+x^{9}\right|}{9}+C$$
