## anonymous 5 years ago what is the integral of 1/x^2

1. anonymous

i dont even think its possible -.-

2. anonymous

-1/x+C

3. anonymous

that makes sense...

4. anonymous

lol, you can let u = x and du = dx and solve :) loki will show you the steps ^_^

5. anonymous

You just look at it like this,$\int\limits_{}{}\frac{1}{x^2}dx=\int\limits_{}{}x^{-2}dx=\frac{x^{-1}}{-1}+c=-x^{-1}+c=-\frac{1}{x}+c$

6. anonymous

integral x^(-2) Just add one to the exponent and divide by the result

7. anonymous

Yep.

8. anonymous

what method is that?

9. anonymous

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.

10. anonymous

?????????

11. anonymous

Usual method...just the same you're used to when you integrate something like$\int\limits_{}{}x dx$

12. anonymous

$\int\limits_{}{}x^ndx=\frac{1}{n+1}x^{n+1}+c$where n is any (real) number.

13. anonymous

Here you started with n=-2.

14. anonymous

Damn I overcomplicated that

15. anonymous

Hehe...slightly ;)

16. anonymous

slightly... dudee i used u sub, uv vdu sub, pfd, random stuff i could come up with.. i WAYYYY overcomplicated it :P

17. anonymous

lol, you should calm down.

18. anonymous

probably, so the integral of lnx/x^2 is lnx/x - 1/x +c

19. anonymous

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.

20. anonymous

I get a minus outside of your lnx/x

21. anonymous

you say that to everyone who askes for conformation lol and my bad, i left it back in uv vdu sub -.-

22. anonymous

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'.