Improper integral

[learningsumthing]

Hey guys,
I'm trying to figure out how wolfram solves the following improper integral
from x=1 to infinity dx/(x(2x+5))

When I solve it as an indefinite, I get the same answer as wolfram (1/5)(ln|x| - ln|2x+5|). However, once I try to integrate with the 1 and infinity, I get (1/5)ln|7| as my final answer, as opposed to Wolfram's (1/5)ln|7/2|

Can someone explain this to me please? I'm so confused.

Thank you so much.

 

[galactus]

$\displaystyle \int\frac{1}{x(2x+5)}dx=\frac{ln(x)}{5}-\frac{ln(2x+5)}{5}$

$\displaystyle \lim_{x\to \infty}\left[\frac{ln(x)}{5}-\frac{ln(2x+5)}{5}\right]=\frac{-ln(2)}{5}$

\(\displaystyle \left \frac{ln(x)}{5}-\frac{ln(2x+5)}{5}\right|_{x=1}=\frac{-ln(7)}{5}\)

$\displaystyle \frac{-ln(2)}{5}-(\frac{-ln(7)}{5})=\frac{ln(7)}{5}-\frac{ln(2)}{5}\overbrace{=}^{\text{ln(a)-ln(b)=ln(a/b)}}=\frac{ln(7/2)}{5}$