<s>[INST] Prove that $\int_0^1 x(1+x^2)^n dx = \dfrac{1}{2n+2}$ $\int_0^1 x(1+x^2)^n dx = \dfrac{1}{2n+2}$
How is this calculated?
This is an interesting case because as a consequence of this, $\int_0^1 n^2x(1+x^2)^n dx$ converges to infinity with n to infinity, but the integrant converges point wise to 0.
(see Rudin "Real and complex Analysis" example 7.6.)
 [/INST] We use substitution rule with $1 - x^2 = t$. We have 
\begin{align}
   &dt = -2x dx \nonumber \\
   &x dx = - \frac{1}{2} dt \nonumber
\end{align}
\begin{align}
  &x = 0, \quad t = 1 - 0 = 1 \nonumber \\
  &x = 1, \quad t = 0 \nonumber
\end{align}
Pulling it all into one single expression.
$$
\int_{0=x}^{1=x} x(1-x^2)^n dx = \int_{0=t}^{1=t} \cdot \left( - \frac{1}{2} \right) t^n dt = \frac{1}{2} \int_{0}^{1} t^n dt = \frac{1}{2} \cdot \frac{t^{n+1}}{n+1}  \Bigg\rvert^1_0 = \frac{1}{2} \frac{1}{n+1} $$
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