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

1. AravindG

quotient rule may be

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

3. anonymous

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.

4. anonymous

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

5. anonymous

oh right split it into 2 fractions first

$=\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$

7. anonymous

This kind of idea comes from experience I guess?

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

9. anonymous

ok, thx for your help :)

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

11. TuringTest

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

12. anonymous

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

13. TuringTest

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

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

15. TuringTest

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

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

17. anonymous

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

18. TuringTest

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.

19. TuringTest

the degree of the numerator*

20. anonymous

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.