Esplain please..very confused...(calculus)?

Question

how does : xlnx(2x+1) - x - (1/2)ln(2x+1) + C

become : (1/2)(2x+1)ln(2x+1)-x+C

??????

the original problem is intigration by parts: integrate ln(2x+1)dx

Answer 1

xlnx(2x+1) - x - (1/2)ln(2x+1) + C

= xlnx(2x+1) - (1/2)ln(2x+1) - x + C

= (x - (1/2)) ln(2x+1) - x + C

= ((2/2)x - (1/2)) ln(2x+1) - x + C

= (1/2)(2x - 1) ln(2x+1) - x + C

and that's not what you wanted. Could it be, that the first line is:

xlnx(2x+1) - x - (1/2)ln((2x+1)^(-1)) + C?

Then ln(z^-1)=-ln(z), and you would get your result...

Additional Details: If the first line reads:

x ln (2x+1) - x + 1/2 ln(2x+1) + c

You just need to change the sign in front of the summands given, follow the lines, just substitute + by - and you'll get

(1/2)(2x + 1) ln(2x+1) - x + C

(the -1 will change into a +1).

Answer 2

By parts:  

   d ( u v ) = u dv + v du

so uv = int u dv + int v du

Here take u=x and v = ln(2x+1)

Any way,  to clean up the answer,

x ln (2x+1) - 1/2 ln(2x+1)    factorises to 1/2 (2x -1) ln (2x+1)

Must be a typo somewhere.