Question

Which method can help me to do the integral of (x^2+1)/(1-x^2)?

Answer 1

quotient rule may be

Answer 2

write x^2+1 as x^2-1+2, make two basic integrals and solve

Answer 3

Because 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.

Answer 4

ha i was just going to suggest partial fractions
1-x^2 = (1-x)(1+x)

Answer 5

oh right split it into 2 fractions first

Answer 6

\[=\int\limits_{?}^{?}(x^2-1)/(1-x^2)dx+\int\limits_{?}^{?}2/(1-x^2)=-\int\limits_{?}^{?}dx+2\int\limits_{}^{?}dx/(1-x^2)\]\[=-x+\ln \left| (1+x)/(1-x) \right|+c\]

Answer 7

This kind of idea comes from experience I guess?

Answer 8

yes, it takes some practice to get the routine, but consider it very easy integral...:D

Answer 9

ok, thx for your help :)

Answer 10

if you've got some more problems with integrals hit me now :D

Answer 11

Do you want to see it with partial fractions?
it's not very hard

Answer 12

Yes, might be a nice to see a different approach.

Answer 13

well 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\[\frac{1+x^2}{1-x^2}=\frac{-x^2-1}{x^2-1}=-1-\frac2{x^2-1}\]now doing partial fractions on the second term\[\frac{-2}{x^2-1}=\frac A{x-1}+\frac B{x+1}\]\[A(x+1)+B(x-1)=-2\]\[A=-1,B=1\]so we wind up with\[\int-1+\frac1{x-1}-\frac1{x+1}dx\]again

Answer 14

well, that's nice but you would you make that second part complicated when you know that it's a basic integral: \[\int\limits_{?}^{?}dx/(1-x^2)=(1/2)\ln \left| (1+x)/(1-x)\right|\]

Answer 15

I didn't know that form, but if they are supposed to use partial fractions in the problem they must do it here.

Answer 16

OK, I didn't know that they must do it like that :D

Answer 17

Can I do this partial fractions stuff also without changing x^2+1 to x^2-1+2?

Answer 18

the 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.

Answer 19

the degree of the numerator*

Answer 20

Ah, 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.