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?