what is the integral of 1/x^2

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what is the integral of 1/x^2

Mathematics
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i dont even think its possible -.-
-1/x+C
that makes sense...

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lol, you can let u = x and du = dx and solve :) loki will show you the steps ^_^
You just look at it like this,\[\int\limits_{}{}\frac{1}{x^2}dx=\int\limits_{}{}x^{-2}dx=\frac{x^{-1}}{-1}+c=-x^{-1}+c=-\frac{1}{x}+c\]
integral x^(-2) Just add one to the exponent and divide by the result
Yep.
what method is that?
Same deal. Don't be put off by the negative sign. Also, if you ever have a number that's not an integer, don't be put off by that either: add 1 and divide by the result.
?????????
Usual method...just the same you're used to when you integrate something like\[\int\limits_{}{}x dx\]
\[\int\limits_{}{}x^ndx=\frac{1}{n+1}x^{n+1}+c\]where n is any (real) number.
Here you started with n=-2.
Damn I overcomplicated that
Hehe...slightly ;)
slightly... dudee i used u sub, uv vdu sub, pfd, random stuff i could come up with.. i WAYYYY overcomplicated it :P
lol, you should calm down.
probably, so the integral of lnx/x^2 is lnx/x - 1/x +c
You can check by taking the derivative of your answer. If it's right, it should equal the integrand (the expression inside the integral). It's the Fundamental Theorem of the Calculus.
I get a minus outside of your lnx/x
you say that to everyone who askes for conformation lol and my bad, i left it back in uv vdu sub -.-
You should learn a method called, 'Tabular Integration by Parts'. If you understand the theory, it's good to know this method because you can punch out IBP problems with a lot less hassle. There's a good explanation about it here: http://en.wikipedia.org/wiki/Integration_by_parts under, ironically, 'Tabular Integration by Parts'.

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