I think I have gone wrong with the integration of (u-1) somehow because according to the book's answer this is supposed to come out at 7u7 (before substituting the original value of u back in).
I can't see how what I've done is wrong though because I've followed the principle: ∫(ax+b)ndx=a(n+1)1(ax+b)n+1
I was wondering if anyone could help me with this problem which I can't work out how to solve.
∫4x(2x+1)5dxu=2x+1
This is what I have done:
du=2dx dx=21du
∫4x(u)521du
21∫4x(u)5du
But I am not sure how to get rid of the 4x and proceed from here. Any help would be much appreciated!
Are you being asked to use substitution here? Personally I would just do it directly (sorry to type up a solution, but, as your question has been answered I thought I would include this as a bonus if you're interested)