integral of dx/x^2+2x-3

I used partial fractions and got 1/3ln(x-1)-1/3ln(x+2) but I ran it through Symbolab just to check and they're getting 1/4 instead of one third.

Can somebody who's good at partial fractions take a look and check this for me?

Question

integral of dx/x^2+2x-3

I used partial fractions and got 1/3ln(x-1)-1/3ln(x+2) but I ran it through Symbolab just to check and they're getting 1/4 instead of one third.

Can somebody who's good at partial fractions take a look and check this for me?

xapproachesinfinity · Answer

int(1/x^2+2x-3)dx this  is your integral?

xapproachesinfinity · Answer

just a correction by the way the factored form is $(x-1)(x+3)$ not what you did

xapproachesinfinity · Answer

most likely that caused you to lose some important numbers
and you got that 1/3

theoreo · Answer

Oh jeez, I don't know what's wrong with me today. I shouldn't be doing calculus if I'm screwing up simple factoring like that. Thanks man. I think I need to take a break from numbers for a while.

xapproachesinfinity · Answer

Eh that's really okay! i do some mistakes like those as well
just check back if everything is right! you won't have this in exams though! so you need to be cautious

xapproachesinfinity · Answer

I think the rest is okay once you see this mistake!
good luck^_^

theoreo · Answer

Yeah I can take it from here. Still gonna give it a rest for a minute though.

xapproachesinfinity · Answer

go ahead^_^