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anonymous
 4 years ago
Which method can help me to do the integral of (x^2+1)/(1x^2)?
anonymous
 4 years ago
Which method can help me to do the integral of (x^2+1)/(1x^2)?

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nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1write x^2+1 as x^21+2, make two basic integrals and solve

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because I just learned in school to do a partial fraction decomposition but this somehow does not work out very well. I´ll use your tip nenadmatematika.

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0ha i was just going to suggest partial fractions 1x^2 = (1x)(1+x)

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0oh right split it into 2 fractions first

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1\[=\int\limits_{?}^{?}(x^21)/(1x^2)dx+\int\limits_{?}^{?}2/(1x^2)=\int\limits_{?}^{?}dx+2\int\limits_{}^{?}dx/(1x^2)\]\[=x+\ln \left (1+x)/(1x) \right+c\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This kind of idea comes from experience I guess?

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1yes, it takes some practice to get the routine, but consider it very easy integral...:D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, thx for your help :)

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1if you've got some more problems with integrals hit me now :D

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2Do you want to see it with partial fractions? it's not very hard

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, might be a nice to see a different approach.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2well actually, I guess nenad did use them it do the second integral (or at least I don't see how he did without them) another way to get to that point it to rearrange the fraction so you can perform long division\[\frac{1+x^2}{1x^2}=\frac{x^21}{x^21}=1\frac2{x^21}\]now doing partial fractions on the second term\[\frac{2}{x^21}=\frac A{x1}+\frac B{x+1}\]\[A(x+1)+B(x1)=2\]\[A=1,B=1\]so we wind up with\[\int1+\frac1{x1}\frac1{x+1}dx\]again

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1well, that's nice but you would you make that second part complicated when you know that it's a basic integral: \[\int\limits_{?}^{?}dx/(1x^2)=(1/2)\ln \left (1+x)/(1x)\right\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2I didn't know that form, but if they are supposed to use partial fractions in the problem they must do it here.

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.1OK, I didn't know that they must do it like that :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can I do this partial fractions stuff also without changing x^2+1 to x^21+2?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2the order of the numerator has to be smaller than that in the denominator to do partial fractions, so you need to make that happen somehow long division is a very standard procedure in these cases, but you can also do what nenad did frequently.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2the degree of the numerator*

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, Thank you. We only talked in school about the cases where the degree is higher or lower but never about what it means when they have the same degree. :( My teacher sucks... Thank you TuringTest for your help and explanation.
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