Improper integral from 1 to infinity

[yeny]

View attachment 8443

Hello everyone,

I am stuck on this homework problem. I got up to (ln (b / (b+1) - ln 1 / (1+1) ) but I'm not sure how to go to the red boxed step where they have (1 - 1 / (b+1) )

if anyone can figure it out Id really appreciate it.

thank you very much.

 

[joypav]

$\frac{b}{b+1} = \frac{b+1-1}{b+1} = \frac{b+1}{b+1} + \frac{-1}{b+1} = 1 - \frac{1}{b+1}$

 

[yeny]

THANK YOU soo much for your help and time. Really appreciate it!