M Mattamz Member Joined Aug 13, 2005 Messages 64 Gender Male HSC 2007 Jun 24, 2007 #2 two application of intergration by parts first time: u=(Inx)^2 du= (2Inx)/x v=x dv=dx I= x(Inx)^2 - Int 2(Inx) dx then r = Inx dr = 1/x dx s = x ds = dx I = x(Inx)^2 - 2[Inx/x - Int 1.dx] =x(Inx)^2 - 2xInx + 2x + C EDIT: forgot the constant '2' on the x term Last edited: Jun 24, 2007
two application of intergration by parts first time: u=(Inx)^2 du= (2Inx)/x v=x dv=dx I= x(Inx)^2 - Int 2(Inx) dx then r = Inx dr = 1/x dx s = x ds = dx I = x(Inx)^2 - 2[Inx/x - Int 1.dx] =x(Inx)^2 - 2xInx + 2x + C EDIT: forgot the constant '2' on the x term
R roadrage75 Member Joined Feb 20, 2007 Messages 107 Gender Male HSC 2007 Jun 24, 2007 #3 I((logx)^2) = x(logx)^2 -I(((2x(logx)/x) = x(logx)^2 -I(2logx) = x(logx)^2 -((2xlogx) -I(2x/x))) = x(logx)^2 -2xlogx + 2x so to integrate call (logx)^2 u, and 1 v' then, as evident in line 2 you have to do it again, this same call logx u and 2 v'
I((logx)^2) = x(logx)^2 -I(((2x(logx)/x) = x(logx)^2 -I(2logx) = x(logx)^2 -((2xlogx) -I(2x/x))) = x(logx)^2 -2xlogx + 2x so to integrate call (logx)^2 u, and 1 v' then, as evident in line 2 you have to do it again, this same call logx u and 2 v'
