# 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.

• Chain rule plus the fundamental theorem of calculus. – Daniel Fischer Sep 17 '14 at 17:55
• Why not integration by parts again ? – Claude Leibovici Sep 17 '14 at 17:58
• $(f(x)^{n+1})'=(n+1)f(x)^nf'(x)$ by the chain rule. – k5f Sep 17 '14 at 18:06
• This identity can easily be derived if you know integration by substitution. – Alice Ryhl Sep 17 '14 at 20:05

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.

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.

By the chain rule we have

$$(g\circ f)'=(g'\circ f)\times f'$$ Now what we get if we take $g(x)=x^n$?