Confused about a certain an integral

Question

I'm studying calculus at school and also via Khan academy. The current course on Khan academy is using Partial Fraction Expansion and I understand how it works, however, for the solution for one of the practice problems

$$\int\frac{-2}{2x-1}\,dx = -\ln |2x-1|$$

while

$$\int\frac{3}{x+4}\,dx = 3\ln |x+4|$$

The later makes sense to me but how come $\int\frac{-2}{2x-1}\,dx = -\ln |2x-1|$ rather than $-2\ln|2x+1|$?

I was very confused so I checked the Khan academy facebook page and also the comments on that section. I've searched as usually a yahoo questions or quora page will come up but nothing as explained to me where the $2$ in $-2$ goes.

Answer 1

1. Substitute $u=2x-1$ and $\text{d}u=2\space\text{d}x$: $$\int\frac{-2}{2x-1}\space\text{d}x=-2\int\frac{1}{2x-1}\space\text{d}x=\frac{1}{2}\cdot(-2)\int\frac{1}{u}\space\text{d}u=-\int\frac{1}{u}\space\text{d}u=\text{C}-\ln\left|u\right|$$
2. Substitute $s=x+4$ and $\text{d}s=\text{d}x$: $$\int\frac{3}{x+4}\space\text{d}x=3\int\frac{1}{x+4}\space\text{d}x=3\int\frac{1}{s}\space\text{d}s=3\ln\left|s\right|+\text{C}$$

Answer 2

Write: $\frac{-2}{2x+1}=-\frac{2}{2x+1}$ and consider that an integral function of $-f(x)$ is minus an integral function of $f(x)$.