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?
int(1/x^2+2x-3)dx this is your integral?
just a correction by the way the factored form is \((x-1)(x+3)\) not what you did
most likely that caused you to lose some important numbers and you got that 1/3
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.
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
I think the rest is okay once you see this mistake! good luck^_^
Yeah I can take it from here. Still gonna give it a rest for a minute though.
go ahead^_^
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