<s>[INST] evaluation of $\int\frac{x^5}{x^5+x+1}dx$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx$
$\bf{My\; Try::}$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx = \int\frac{\left(x^5+x+1\right)-(x+1)}{x^5+x+1}dx = x-\int\frac{x+1}{x^5+x+1}dx$
Now Let $\displaystyle I = \int\frac{x+1}{x^5+x+1}dx = \int \frac{x+1}{(x^2+x+1)\cdot (x^3-x^2+1)}$
Now I Did not understand how can i solve after that
Help Required
Thanks
 [/INST] Note that\begin{align}
&\frac{x^5}{x^5+x+1} 
=1-\frac{2x+3}{7(x^2+x+1)}+
\frac{2x^2-x-4}{7(x^3-x^2+1)}\\
 =& \ 1-\frac{2x+1}{7(x^2+x+1)}-\frac{2}{7(x^2+x+1)} +
\frac{2(3x-2x)}{21(x^3-x^2+1)} +
\frac{x-12}{21(x^3-x^2+1)}
\end{align}
Then
$$\int \frac{x^5}{x^5+x+1} dx
=x -\frac1{21}\ln\frac{(x^2+x+1)^3}{(x^3-x^2+1)^2}
-\frac4{7\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3}+\frac1{21}K
$$
where
$$K = \int \frac{x-12}{x^3-x^2+1}dx$$
$x^3-x^2+1= (x-a)(x^2-\frac x{a^2}-\frac1a) $ has a single real root $a=-0.7549$, or
$$a =\frac13\bigg( 1-\sqrt[3]{\frac{25+3\sqrt{69}}2}-\sqrt[3]{\frac{25-3\sqrt{69}}2}\bigg)
$$
which allows $K$ to be integrated below with partial fractions
\begin{align}
K= &\int\frac{x-12}{(x-a)(x^2-\frac x{a^2}-\frac1a)}dx\\
=& \ \frac1{a^3-2}\int
 \frac{a(a-12)}{x-a}+ \frac{a(12-a)(2x-\frac1{a^2})}{2(x^2-\frac x{a^2}-\frac1a)}+ \frac{12a^2-\frac6a-\frac32}{x^2-\frac x{a^2}-\frac1a}\ dx\\
=& \ \frac1{a^3-2}\bigg(\frac{a(a-12)}{2}\ln\frac{(x-a)^2}{x^2-\frac x{a^2}-\frac1a}
+\frac {a^2-16a-4}{\sqrt{-4a^3-1}}\tan^{-1}\frac{2a^2x-1}{\sqrt{-4a^3-1}}\bigg)
\end{align}
</s>