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: 
if you write $f(x) = 1-x^2$, your integral can be rewriten as 

$$\int\_0^1 nx(1-x^2)^n = \int\_0^1 -\frac{n}{2} f'(x)f(x)^n dx $$

$$ = \int\_0^1 -\frac{n}{2(n+1)} \left( f(x)^{n+1} \right)' dx$$

$$ = \left[-\frac{n}{2(n+1)} f^{n+1}(x)\right]\_0^1$$
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 $\frac{n}{2(n+1)}$