<s>[INST] Finding $\int_0^1{\frac{x^4(1-x)^4}{1+x^2}}dx$ The question I am working on: Evaluate 
$$\frac{1}{2} \int^1_0{x^4 (1-x)^4 } dx \le \int^1_0{\frac{x^4 (1-x)^4}{1+x^2}} dx \le \int^1_0{x^4 (1-x)^4 } dx$$
So using integration by parts to solve: 
(letting $u=(1-x)^4$ and $dv=x^4$)
$$\int{x^4 (1-x)^ 4} dx = \frac{4}{5}x^5(x-1) - \frac{4}{5}\left(\frac{x^7}{7} - \frac{x^6}{6}\right)+c$$
Is it correct so far? If so ... 
$$\int^1_0{ x^4 (1-x)^4 } dx = \frac{4}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac2{105}$$
$$\frac{1}{2} \int^1_0{ x^4 (1-x)^4 } dx = \frac{2}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac1{105}$$
But 
$$\int\frac{{x^4(1-x)^4}}{1+x^2} dx = ??$$

Since I found $\int{{x^4(1-x)^4}} dx$. I thought of integration by parts, letting $u=\frac{1}{1+x^2}$, $dv = x^4(1-x)^4$. But I will get a very complicated $v$ to integrate later? Same if I did it the other way around? 
 [/INST] Using the binomial theorem and then long division,
$$
\frac{x^4(1-x)^4}{1+x^2}=
\frac{x^4(x^4-4x^3+6x^2-4x+1)}{1+x^2}=
x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}
$$
so that
$$
\int_0^1
\frac{x^4(1-x)^4}{1+x^2}dx=
\frac17-\frac46+\frac55-\frac43+\frac41-4\arctan\frac\pi4
=\frac{22}7-\pi\approx\frac1{790.~833~125~927~563}
$$
There's also a problem with your integration by parts since
$$
\eqalign{
  u &=& (1-x)^4=(x-1)^4  &\qquad&   v &=& x^4 \\\\
 du &=& 4(x-1)^3 \, dx   &\qquad&  dv &=& \frac15 x^5 dx
}
$$
so
$$
\eqalign{
I_1
&= \int_0^1 x^4 (1-x)^4 dx \\
&= \int u\,dv = uv - \int v\,du \\
&= \left[ \frac45 x^5 (1-x)^4 \right]_0^1
 + \frac45 \int_0^1 x^5 (1-x)^3 dx \\
&= \frac45 \int_0^1 x^5 (1-x)^3 dx \\
&= \frac45\cdot\frac36 \int_0^1 x^6 (1-x)^2 dx \\
&= \frac45\cdot\frac36\cdot\frac27 \int_0^1 x^7 (1-x) dx \\
&= \frac45\cdot\frac36\cdot\frac27\cdot\frac18 \int_0^1 x^8 dx \\
&= \frac45\cdot\frac36\cdot\frac27\cdot\frac18\cdot\frac19 = \frac1{630}
}
$$
where we had to apply integration by parts repeatedly using
$$u=(1-x)^n,~dv=x^m\,dx$$
$$du=-n(1-x)^{n-1}\,dx,~v=\frac{x^{m+1}}{m+1}\implies$$
$$\int x^m(1-x)^ndx=\frac{x^n(1-x)^{m+1}}{m+1}
+\frac{n}{m+1}\int x^{m+1}(1-x)^{n-1}dx$$
until the powers of $(1-x)$ went away. In fact,
we just made a special case of the calculation
$$
\int_0^1 x^a (1-x)^b = B(a+1,b+1) = \frac{a!~b!}{(a+b+1)!}
$$
of the well-known Beta function for $a=b=4$
(our method works for $a,b\in\mathbb{N}$ and,
with an adaptation of the above formula
using the Gamma function, also for real $a,b\ge0$).
Now we can see that the reciprocal of the central integral
is certainly beween $630$ and $1260$.
</s>