Induction proof for integrals 
I am having issues with proving the second step, and have been told that it would be best to approach this question by induction. I have manages to get this far:
$$\left(\int_{a}^b g(x)f(x)=\frac{(-1)^{k}}{c^{k}}\int_{a}^b g(x)f^{(k)}(x)\right)\times\frac{-1}{c}$$
$$\frac{-1}{c}\int_{a}^b g(x)f(x)=\frac{(-1)^{k+1}}{c^{k+1}}\int_{a}^b g(x)f^{(k)}(x)$$
$$-\int_{a}^b g(x)f(x)=\frac{c(-1)^{k+1}}{c^{k+1}}\int_{a}^b g(x)f^{(k)}(x)$$
$$-\int_{a}^b g(x)f(x)=\frac{(-1)^{k+1}}{c^{k+1}}\int_{a}^b cg(x)f^{(k)}(x)$$
$$-\int_{a}^b g(x)f(x)=\frac{(-1)^{k+1}}{c^{k+1}}\int_{a}^b g'(x)f^{(k)}(x)$$
From here I have no idea how to produce $f^{(k+1)}(x)$ inside the integral, or whether this was the right way to approach solving this problem. Any pointers with how to proceed would be awesome!
Cheers  
 A: Suppose it is true for $n=k$. Accordingly, we assume that the following holds
$$\int_a^b {g(x)f(x)dx = {{{{( - 1)}^k}} \over {{c^k}}}\int_a^b {g(x){f^{(k)}}(x)dx} } $$
Now let's go to see what happens about $n=k+1$. We use integration by parts for the right hand side of the above equation to get
$$\eqalign{
  & \int_a^b {g(x)f(x)dx = \left[ {{{g(x)} \over c}{f^{(k)}}(x)} \right]_a^b - {{{{( - 1)}^k}} \over {{c^k}}}\int_a^b {{{g(x)} \over c}{f^{(k + 1)}}(x)dx} }   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0 + {{{{( - 1)}^{k + 1}}} \over {{c^{k + 1}}}}\int_a^b {g(x){f^{(k + 1)}}(x)dx}   \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{{{( - 1)}^{k + 1}}} \over {{c^{k + 1}}}}\int_a^b {g(x){f^{(k + 1)}}(x)dx}  \cr} $$
and hence it is also valid for $n=k+1$. Cheers! :D 

Edit 
Integration by parts formula is
$$\eqalign{
  & V(x) = \int_k^x {v(t)dt}  + A  \cr 
  & \int_a^b {u(x)v(x)dx}  = \left[ {u(x)V(x)} \right]_a^b - \int_a^b {V(x)u'(x)dx}  \cr} $$
where $V(x)$ is an anti-derivative (or primitive function) of $v(x)$. In your problem we should choose
$$\eqalign{
  & u(x) = {f^{(k)}}(x)  \cr 
  & v(x) = g(x)  \cr 
  & V(x) = \int_k^x {g(t)dt}  + A = \left[ {{{g(t)} \over c}} \right]_k^x + A = {{g(x)} \over c} + \left( { - {{g(k)} \over c} + A} \right) \cr} $$
but since $A$ is arbitrary we can choose $A = {{g(k)} \over c}$ and hence
$$V(x) = {{g(x)} \over c}$$
This will help you to understand the integration by parts done in the above solution.
