Finding Integrals?

Question

Ok, so I'm starting Cal. 2 and I'm already lost

Given the integral ln(2x+1)dx, how do I get the answer, which is: 1/2(2x+1)ln(2x+1) -x +c?

u = ln 2x+1
du = 2/2x+1 dx
dv= dx
v= x

integral ln (2x+1)= x ln (2x+1) - integral 2x/(2x+1)dx
= x ln (2x+1) - integral 1- 1/(2x+1)dx
= x ln (2x+1) - x + ln |2x+1| + c
then I'm lost....

Thank you!

Answer 1

integral ln (2x+1)= x ln (2x+1) - integral 2x/(2x+1)dx ... ok
= x ln (2x+1) - integral 1- 1/(2x+1)dx

wait, see the 3rd term:
= x ln (2x+1) - x + (1/2)ln |2x+1| + c
= (x+1/2) ln (2x+1) - x + c
= 1/2(2x+1)ln (2x+1) - x + c

Note:
integral 1/(2x+1) dx = 1/2 ln|2x+1|+k
..... this is where the mistake occurred....


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Answer 2

I think this gets easier if we first solve Integral ln(u) du. By parts, similarly to what you did, it follows that

Integral ln(u) du = u ln(u) - Integral u 1/u du = u ln(u) - Integral du = u ln(u) - u + C.

Now, put 2x + 1 = u, so that x = (u -1)/2 and dx = du/2. Then

Integral ln(2x+1)dx = Integral ln(u) u/2 du = 1/2( uln(u) - u) + C = 1/2((2x +1) ln(2x+1) - 2x -1) + C