Question: > 
> Evaluate the integral $\int\_0^1 nx(1-x^2)^n \, dx$ ($n$ is constant here)
> 
> 
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I actually have the solution which is: 

$$\int\_0^1 nx(1-x^2)^n = \left( \frac{-n}{2(n+1)} (1-x^2)^{n+1}\right)|\_0^1 = \frac{n}{2(n+1)}$$

It's unclear to me this was calculated. I'd be glad for reasoning.

Thanks.

Answer: 
We want to integrate $f(x)=nx(1-x^2)^n$ 

Let $u=1-x^2$ and thus $du=-2xdx$

$\int{f(x)dx}=\frac{-1}{2}\int{n\cdot u^n (-2x)dx}=\frac{-1}{2}\int{nu^ndu}=-\frac{nu^{n+1}}{2(n+1)}+c=-\frac{n(1-x^2)^{n+1}}{2(n+1)}+c$

Integrating from 0 to 1, we get $\frac{n}{2(n+1)}$
Hint:
$$\text{If } u = (1 - x^2)^k,\ du = k(1-x^2)^{k-1}(-2x) dx= -2kx(1-x^2)^{k-1}dx$$
Then
$$\int nx(1-x^2)^n dx = -\frac{n}{2(n+1)}\int (n+1)(-2x)(1-x^2)^n dx$$
$$= -\frac{n}{2(n+1)}\int (1-x^2)^n\left((n+1)(-2x)\ dx\right)$$