Calculate Indefinite Integral $\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx$ The following integration is given by Wolfram Alpha
$$\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}.$$
My question is: what is the best (meaning least work), method to achieve this result by hand ? There are two approaches I see, partial fractions, maybe setting $y=1+x$, but this is still alot of work. Or perform successive integration by parts. Both still involve a lot of calculation. The compact form of the answer makes me hope there is a nice way to achieve it. Guessing the ultimate form and then calculating the derivative is, to me, a less desirable method.
 A: 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}
A: 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.
A: $$\int \frac{x^5 \left(1-x^6\right)}{(x+1)^{18}}\,dx=\frac{P(x)}{(x+1)^{17}}$$
$$\frac{x^5 \left(1-x^6\right)}{(x+1)^{18}}=\frac{(x+1) P'(x)-17 P(x)}{(x+1)^{18}}$$ So, $P(x)$ is a polynomial of degree $11$.
Let
$$P(x)=\sum_{n=0}^{11} a_n\,x^n$$ Expand and group powers to get
$$0=(a_1-17 a_0)+(2 a_2-16 a_1) x+(3 a_3-15 a_2) x^2+(4 a_4-14 a_3) x^3+$$ $$(5
   a_5-13 a_4) x^4+(-12 a_5+6 a_6-1) x^5+(7 a_7-11 a_6) x^6+(8 a_8-10 a_7)
   x^7+$$ $$(9 a_9-9 a_8) x^8+(10 a_{10}-8 a_9) x^9+(11 a_{11}-7 a_{10}) x^{10}+(1-6
   a_{11}) x^{11}$$ which is simple.
If I not wrong the coefficients are
$$\left\{0,0,0,0,0,0,\frac{1}{6},\frac{11}{42},\frac{55}{168},\frac{55}{168},\frac{11}{42},\frac{1}{6}\right\}$$ and factoring
$$P(x)=\frac{1}{168} x^6 (x+1) \left(28 x^4+16 x^3+39 x^2+16 x+28\right)$$
A: Define  $F_1(x)=\int f(x) dx$. $F_{n+1}(x)=\int F_n(x) dx$ and note the following (using integration by parts) :
\begin{align*}
\int x^n f(x)dx &=x^n F_1(x)-\int nx^{n-1}F_1(x)dx\\&=x^n F_1(x)-nx^{n-1}F_2(x)+n(n-1)\int x^{n-2} F_3(x)dx
\end{align*}
Note that the pattern on RHS will end when $x^n$ reduces to a constant (because $x^n$ is being differentiated and $F$ is being integrated.)
Based on this observation, let's try to write the given integral in two parts:
$\int \frac{x^5}{(1+x)^{18}}dx=(x^5)\bigg(\frac{(1+x)^{-17}}{-17}\bigg)-(5x^4)\bigg(\frac{(1+x)^{-16}}{(-17)(-16)}\bigg)+(20x^3)\bigg(\frac{(1+x)^{-15}}{(-17)(-16)(-15)}\bigg)-(60x^2)\bigg(\frac{(1+x)^{-14}}{(-17)(-16)(-15)(-14)}\bigg)+(120x)\bigg(\frac{(1+x)^{-14}}{(-17)(-16)(-15)(-14)}\bigg)-(120)\bigg(\frac{(1+x)^{-13}}{(-17)(-16)(-15)(-14)(-13)}\bigg)$
and the second part (that is the integral $\int \frac{x^5.x^6}{(1+x)^{18}}dx$) can be written similarly. With that we are done.
