<s>[INST] Computing $\int \frac{7x}{(2x+1)} dx$ I'm trying to compute the following integral $$\int \frac{7x}{(2x+1)} dx$$
Unfortunately Wolfram Alpha gives me a different result, but other integral calculators say that my result is correct. So where is my error:
$$\int \frac{7x}{(2x+1)} dx = \frac{7}{2}\int \frac{2x}{(2x+1)} dx$$
$$\frac{7}{2}\int \frac{2x+1}{(2x+1)} - \frac{1}{(2x+1)} dx = \frac{7}{2}\int 1 dx - \int \frac{1}{(2x+1)} dx$$
Let $u=2x+1$
$$=\frac{7}{2}(x-\frac{1}{2}\int\frac{1}{u}du)=\frac{7x}{2}-\frac{7}{4}\ln(|2x+1|)+C$$
Wolfram Alpha tells me that it is $$\frac{7}{4} (2 x - \ln(2 x + 1) + \underbrace{1}_{different}) + C$$
Why is there an additional $1$?
 [/INST] 
If $y=F(x)$ be a primitive of the function $y=f(x)$ then for every real $k$ the function $y=F(x)+k$ is a primitive also.

The primitive of $\dfrac{7x}{2x+1}$ is $\dfrac{7}{4} (2x -\ln(2 x + 1))$, then for every real $k$ the function $y=\dfrac{7}{4} (2x -\ln(2 x + 1))+k$ is a primitive also.
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