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but in the alternate forms it does absorb the constant :D

right^

turing i need u 1 min

i will give u the link

what are you guys doing?

comparing your answers with wolfram?

sprusty 1/4+constant is still a constant
so your answer and the book answer are fine! :)

yes, but why does wolf add the 1 for 'restricted values' I want to know

\[=\frac{1}{4}(2x+1-\ln|2x+1|)+C\]

\[=\frac{x}{2}+\frac{1}{4}-\frac{1}{4} \ln|2x+1| +C\]

\[=\frac{x}{2}-\frac{1}{4}\ln|2x+1|+c \]

i will look at wolfram's solution one sec

No I don't know what they are talking about honestly lol

either one of those things they have listed is fine though

imperialist seems smart
I will ask him what he think wolfram means