Computing $\int \frac{7x}{(2x+1)} dx$

Question

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 $1$?

Answer 1

Note that $C$ is an arbitrary constant, so both answers are actually correct. One may thus let $k=\frac{7}{4}+C$ and then obtain the same answer you've obtained.

Answer 2

If you look at what Wolfram gives you

$$\frac{7}{4} (2 x - \ln(2 x + 1) +1 )+ C$$

and see that you have an "extra" term, look at it this way: any constant terms that differ from your result may have been "absorbed" into the arbitrary constant.

What you get is an antiderivative, the important thing when you evaluate indefinite integrals is that you can get back to the original integrand when you take the derivative of your result.

Answer 3

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.