Integration By Parts: Need help with a step

[daviddee305]

Integration By Parts: Need help with a step...

Evaluate the integral:

\int ln(2x + 1)dx

I worked it out up until:

Xln(2x + 1) - \int 2x/(2x + 1) dx

Then the next step throws me off. I attached a scan from the solutions manual and circled the part that confused me. Could somebody explain how we went from "2x/(2x + 1)" to "[(2x + 1) - 1]/(2x + 1)"?

I'm sure it's a simple arithmetic step I'm overlooking, which is why I'm reviewing a few weeks before Calc. 2.

 

[Cyosis]

Not much happens really, (2x+1)-1=2x+1-1=2x. As you can see both numerators are the same, the latter is just written in a more useful form.

 

[daviddee305]

thanks... it makes sense now...