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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.

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

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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.

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By the chain rule we have

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

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