calculus - Integration by parts: u substitution case for $ln(2x+1)$

Integrate by parts: $\int \ln (2x + 1) \, dx$

This question has answers here and I was wondering about the last answer to the question which is the way I did it, got a different answer?

https://math.stackexchange.com/a/1598118/372659

Here is the answer,

Is doing u-substitution, and then integration by parts a wrong method here? Because in my textbook the answer there is a ones in the other answers.

Answer

You should find

$$[x\ln (2x+1)]-\int x\frac {2}{2x+1}dx $$

$$=x\ln (2x+1)-\int \frac {2x+1-1}{2x+1}dx $$

$$=x\ln (2x+1)-x+\frac {1}{2}\ln (2x+1) +C$$

$$=\frac {2x+1}{2}\ln (2x+1)-x+C $$

You can also put $u=2x+1$ to get

$$\int \ln (u)\frac {du}{2}=\frac {1}{2}(u\ln (u)-u)+K $$ $$=\frac {2x+1}{2}\ln (2x+1)-x-\frac {1}{2}+K $$

with $C=K-1/2$.