I just did an integral and i'm trying to simplify my final answer:
xln(2x+1)-x+(1/2)(ln(2x+1))+C

Question

mr.math · Answer

Is it $x\ln(2x+1)-x+\frac{1}{2}\ln(2x+1)+C?$

anonymous · Answer

yes

mr.math · Answer

I would recommend that you take $\ln(2x+1)$ as a common factor.

mr.math · Answer

Could you tell me what the original integral was?

anonymous · Answer

the problem is that the answer is $$(1/2)(2x+1)\ln(2x+1)-x+C$$
What happened to the x in xln(2x+1)

anonymous · Answer

The original integral was [int_ln(2x+1)dx]

anonymous · Answer

i set u=ln(2x+1) and dv=dx

mr.math · Answer

Your answer is right then. Here how they did it:
$$x\ln(2x+1)+\frac{1}{2}\ln(2x+1)-x+C=(x+\frac{1}{2})\ln(2x+1)-x+C$$

mr.math · Answer

But if you take $\frac{1}{2}$ as a common factor from the first parentheses, you get
$$\frac{1}{2}(2x+1)\ln (2x+1)-x+C$$

anonymous · Answer

I see. So you multiply by 2 within the parenthesis and divide by 2 to make up for the difference

mr.math · Answer

Exactly!

anonymous · Answer

Ur awesome. Thanks again my friend

mr.math · Answer

You learn fast by the way :)

anonymous · Answer

I disagree...but I'll take the compliment

mr.math · Answer

You have to :P