Question

http://gyazo.com/7919fbc4610aa7647d33100ee01aec29

Answer 1

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

Answer 2

Is that a type-o?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Can you do the test now?

Answer 8

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

Answer 9

You will have to break up the integral

Answer 10

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

Answer 11

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

Answer 12

lol I meant rest above not test

Answer 13

@ggrree any questions?

Answer 14

i think we have a standard formula as well for integrals of this type rather than use partial fractions every time:

\[\int\limits(1/(x^2-a^2))dx =(1/2a)\ln((x-a)/(x+a)) + k\]