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.)
 A: By the change of variable
$$
u=1+x^2, \quad du=2x\: dx,
$$ we rather get
$$
\int_0^1 x(1+x^2)^n dx=\frac12\int_1^2 u^n du=\frac{2^{n+1}-1}{2(n+1)}.
$$

Edit. If you meant $x(1-x^2)^n$ as being the integrand (as suggested by @Pjotr5), the similar change of variable
$$
v=1-x^2, \quad dv=-2x dx,\quad v(0)=1,\quad v(1)=0,
$$ gives

$$
\int_0^1 x(1-x^2)^n dx=\frac12\int_{\color{red}{0}}^{\color{red}{1}} v^n dv=\frac12\cdot\frac{1^{n+1}-0^{n+1}}{n+1}=\frac1{2n+2}.
$$

A: $$\int _{ 0 }^{ 1 } x(1+x^{ 2 })^{ n }dx=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 } (1+x^{ 2 })^{ n }d\left( 1+{ x }^{ 2 } \right) =\frac { 1 }{ 2 } { \left[ \frac { { \left( 1+{ x }^{ 2 } \right)  }^{ n+1 } }{ n+1 }  \right]  }_{ 0 }^{ 1 }=\frac { { 2 }^{ n+1 }-1 }{ 2n+2 } $$
A: 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.
$$
\int_{0=x}^{1=x} x(1-x^2)^n dx = \int_{0=t}^{1=t} \cdot \left( - \frac{1}{2} \right) t^n dt = \frac{1}{2} \int_{0}^{1} t^n dt = \frac{1}{2} \cdot \frac{t^{n+1}}{n+1}  \Bigg\rvert^1_0 = \frac{1}{2} \frac{1}{n+1} $$
