## ggrree 3 years ago http://gyazo.com/7919fbc4610aa7647d33100ee01aec29

1. ggrree

I was able to separate the integrand into partial fractions but then I get stuck.

2. myininaya

Is that a type-o?

3. myininaya

$\int\limits_{0}^{9}\frac{dx}{x^2-6+5}$

4. myininaya

Is it really suppose to be x^2-6+5 and not x^2-6x+5?

5. myininaya

$\frac{1}{x^2-1} \text{ since } -6+5=-1$ $\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$

6. myininaya

$\frac{1}{(x-1)(x+1)}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}$ => 1=x(A+B)+(A-B) A+B=0 ; A-B=1

7. myininaya

Can you do the test now?

8. myininaya

And by the way this is an improper integral meaning the function is not continuous on [0,9]

9. myininaya

You will have to break up the integral

10. myininaya

$\int\limits_{0}^{1}\frac{dx}{x^2-1} +\int\limits_{1}^{9}\frac{dx}{x^2-1}$

11. myininaya

See if both parts converge if so then take the sum in you are done if not then the area/net area is divergent

12. myininaya

lol I meant rest above not test

13. myininaya

@ggrree any questions?

14. apoorvk

i think we have a standard formula as well for integrals of this type rather than use partial fractions every time: $\int\limits(1/(x^2-a^2))dx =(1/2a)\ln((x-a)/(x+a)) + k$

15. apoorvk

|dw:1332635709657:dw|

16. apoorvk

@ggrree

17. ggrree

Yeah, I think I got it. I think... I think it diverges. is that right?

18. ggrree

thanks guys!

19. myininaya

you are right! @ggrree does not converge :)