Question

Integrate the following function: $f' (ax + b) [f (ax + b)]^n$

Answer 1

Answer:

$\displaystyle \int f'(ax+b)\bigl[f(ax+b)\bigr]^n\,dx =\tfrac{1}{a(n+1)}\bigl[f(ax+b)\bigr]^{n+1}$

Step-by-step explanation:

Think about where things like this turn up by differentiating.  This looks almost like the result of using the Chain Rule to differentiate [ f(ax+b) ]^(n+1).

With that in mind...

$\displaystyle \frac{d}{dx} \bigl[f(ax+b)\bigr]^{n+1} = (n+1)\bigl[f(ax+b)\bigr]^n\times\frac{d}{dx}f(ax+b) \\ \\= (n+1)\bigl[f(ax+b)\bigr]^n\times f'(ax+b) \times \frac{d}{dx}(ax+b)\\= a(n+1)f'(ax+b)\bigl[f(ax+b)\bigr]^n \\ \\\Rightarrow \frac{d}{dx} \left(\tfrac{1}{a(n+1)}\bigl[f(ax+b)\bigr]^{n+1}\right) = f'(ax+b)\bigl[f(ax+b)\bigr]^n$

So going backwards...

$\displaystyle \int f'(ax+b)\bigl[f(ax+b)\bigr]^n\,dx =\tfrac{1}{a(n+1)}\bigl[f(ax+b)\bigr]^{n+1}$