Calculate Indefinite Integral $\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx$
Explore the embedded symmetry by denoting
$$f_{n,m}(x)= \frac{x^{n-m}+x^{n+m}}{(1+x)^{2n}}$$
which differentiates as
$$f’_{n,m}(x)
=\frac{(n-m)(x^{-m-1}-x^{m+1}) -2m( x^{-m}-x^{m})-(n+m)(x^{-m+1}-x^{m-1}) }{x^{-n}(1+x)^{2n+2}}
$$
In particular, with $n=8$ and $m=0,1,2$
\begin{align}
f’_{8,0}(x)&= \frac{16(x^7-x^9)}{(1+x)^{18}}\\
f’_{8,1}(x)&= \frac{-2(x^7-x^9) +7(x^6-x^{10})}{(1+x)^{18}}\\
f’_{8,2}(x)&
= \frac{-10(x^7-x^9)-4(x^6-x^{10})+ 6(x^5-x^{11})}{(1+x)^{18}}
\end{align}
which leads to
$$
\frac{x^5-x^{11}}{(1+x)^{18}}
= \frac16 f’_{8,2}(x)+\frac2{21} f’_{8,1}(x)+\frac{13}{112} f’_{8,0}(x)
$$
Integrate to obtain
\begin{align}
\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx
&= \frac16 f_{8,2}(x)+\frac2{21} f_{8,1}(x)+\frac{13}{112} f_{8,0}(x)\\
&= \frac{x^6+x^{10}}{6(1+x)^{16}} + \frac{2(x^7+x^{9})}{21(1+x)^{16}} + \frac{26x^8}{112(1+x)^{16}}\\
&=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}\\
\end{align}
We have the following : \begin{aligned}\frac{x^{5}\left(1-x^{6}\right)}{\left(1+x\right)^{18}}&=\frac{\left(1+x-1\right)^{5}}{\left(1+x\right)^{18}}-\frac{\left(1+x-1\right)^{11}}{\left(1+x\right)^{18}}\\ &=\frac{\sum\limits_{k=0}^{5}{\left(-1\right)^{k}\binom{5}{k}\left(1+x\right)^{5-k}}}{\left(1+x\right)^{18}}-\frac{\sum\limits_{k=0}^{11}{\left(-1\right)^{k}\binom{11}{k}\left(1+x\right)^{11-k}}}{\left(1+x\right)^{18}}\\ \frac{x^{5}\left(1-x^{6}\right)}{\left(1+x\right)^{18}}&=\sum_{k=0}^{5}{\binom{5}{k}\frac{\left(-1\right)^{k}}{\left(1+x\right)^{13+k}}}-\sum_{k=0}^{11}{\binom{11}{k}\frac{\left(-1\right)^{k}}{\left(1+x\right)^{7+k}}}\end{aligned}
Thus : $$ \int{\frac{x^{5}\left(1-x^{6}\right)}{\left(1+x\right)^{18}}\,\mathrm{d}x}=\sum_{k=0}^{5}{\binom{5}{k}\frac{\left(-1\right)^{k+1}}{\left(12+k\right)\left(1+x\right)^{12+k}}}-\sum_{k=0}^{11}{\binom{11}{k}\frac{\left(-1\right)^{k+1}}{\left(6+k\right)\left(1+x\right)^{6+k}}}+C $$
The common denominator is clearly $ \left(1+x\right)^{17} $, so I guess, we could explicit the sums above, simplify and factor, then we'll get the result.