Integration Formula Previously, to integrate functions like $x(x^2+1)^7$ I used integration by parts. Today we were introduced to a new formula in class: $$\int f'(x)f(x)^n dx = \frac{1}{n+1} {f(x)}^{n+1} +c$$ 
I was wondering how and why this works. Any help would be appreciated. 
 A: Hint
Let $u'=f'(x)$ and $v=f(x)^n$; so $u=f(x)$, $v'=n f(x)^{n-1}f'(x)$. Then $$I=\int f'(x)f(x)^n dx =f^{n+1}(x)-n\int f'(x)f(x)^ndx=f(x)^{n+1}-nI$$
I am sure that you can take from here.
A: By the chain rule we have
$$(g\circ f)'=(g'\circ f)\times f'$$
Now what we get if we take $g(x)=x^n$?
A: The reason the formula holds is that for the chain rule:
$$
\left(\frac{1}{n+1}f^{n+1}(x)\right)' =\frac{1}{n+1}(n+1)f^n(x) f'(x) = f^n(x) f'(x)
$$
and this shows your identity by definition of indefinite integral as anti-derivative.
However, applying integration by parts to the same problem we have:
$$
\int f'(x) f^n(x)dx = f(x)f^n(x)-\int f(x)nf^{n-1}(x)f'(x)dx
$$
which is an identity that can be recast precisely as:
$$
\int f'(x) f^n(x) dx =  \frac{1}{n+1}f^{n+1}(x).
$$ 
We can observe that this is a general fact, since the integration by parts rule is just an application of the chain rule itself:
$$
\left(f(x)g(x)\right)' = f'(x)g(x) + f(x)g'(x)\\
f'(x)g(x) = \left(f(x)g(x)\right)'- f(x)g'(x)\\
\int f'(x)g(x) dx = f(x)g(x) - \int f(x)g'(x) dx.
$$
The difference in the two methods above amounts exactly to the difference between the first line and the last one.
