Recent content by sciencefog

I really am confused with the v' =1 i.e. are you saying that the integral of lnx = 1?? :cry: How can that be - as when you differentiate you do not get lnx? If you are not saying that then what is going on = can you break it down for me please as I'm very confused... I thought the...

Why is dv/dx = 1? so here is my attempt at an answer: $2(lnx)^3 u=(lnx)^3, v'=1 uv = x(lnx)^3, vu'=3(lnx)^2 u = (lnx)^2, v'=1 uv= x(lnx)^2, vu'=2lnx so: $udv=x(lnx)^3 -$3(lnx)^2 = x(lnx)^3 - x(lnx)^2 - $2lnx =x(lnx)^3 - x(lnx)^2 - 2(xlnx - x) so...

ok, so can you help me get further than this? maybe pointing me in the direction of the parts bit? I know that: (all this is from the web, I have done this work - but 6 years ago..) integral = $ $udv = uv - $vdu [eg $xe^x.dx, u = x, dv = e^x.dx] so for the expression...

Well I was thinking - but I don't know if its right and I don't know where to find out - hence posting here... :smile: y = 2*(lnx)^3 = 2*[(lnx)(lnx)(lnx)] therefore integral = 2*( xlnx - x * xlnx - x * xlnx - x) +C = 2*(3xlnx - x) + c =6xlnx - x + c Is that right? Thanks very...

Hi, I need some help! :rolleyes: I have a series of maths problems in my physics work... I am trying to integrate a power of a log - such as: y= 3.2875E+004 - 8.9321E+005*ln x + 4.2316E+006*(ln x)^2 - 7.7776E+006*(ln x)^3 + 7.4949E+006*(ln x)^4 - 4.1373E+006*(ln x)^5 +...