Question

Find the integral of 1/(x^2-a^2) dx, where a is not equal to 0. Use partial fraction decomposition. I started the problem, but I'm getting nowhere.

Answer 1

wat u got so far?

Answer 2

did u factor the denominator?

Answer 3

Yes, I factored the denominator using the difference of squares. The factors would be (x+a)(x-a)

Answer 4

ok... so

\(\large \frac{1}{(x+a)(x-a)}=\frac{A}{x+a} +\frac{B}{x-a} \)

now you need to get rewrite the right side so the fractions "can" be combined (common denominators)..

Answer 5

what equation do you get as a result?

Answer 6

\(\large \frac{1}{(x+a)(x-a)}=\frac{A}{x+a} +\frac{B}{x-a} \)

clearing the fractions, we have:

\(\large \frac{1}{(x+a)(x-a)}=\frac{A(x-a)}{(x+a)(x-a)} +\frac{B(x+a)}{(x-a)(x+a)} \)

\(\large 1=A(x-a)+B(x+a) \)

ok so far?

Answer 7

Yes, I've got everything so far.

Answer 8

now just let x=a, solve for B;
then let x=-a, solve for A....

what u got A = ??? , B = ???

Answer 9

That's where I got confused. Why do we let x = a?

Answer 10

to eliminate A so we can solve for B... and we choose specific values of x to make the job easier... notice if we let x=a, then that will eliminate A(x-a) because A(a-a) = 0.
so we end up with the equation:

\(\large 1=A(a-a)+B(a+a) \)
\(\large 1=B(2a) \)

ok so far?

Answer 11

Oh okay, yes I got it.

Answer 12

so, find A and B so we can do the integral...
A =
B =

Answer 13

A= -1/2a and B= 1/2a

Answer 14

great... so we have... (hang on)

Answer 15

\(\large \frac{1}{(x+a)(x-a)}=\frac{A}{x+a} +\frac{B}{x-a} \)
\(\large \frac{1}{(x+a)(x-a)}=\frac{(-1/2a)}{x+a} +\frac{(1/2a)}{x-a}=\frac{-1}{2a(x+a)} +\frac{1}{2a(x-a)}=\frac{1}{2a(x-a)}-\frac{1}{2a(x+a)} \)

ok???

Answer 16

okay, got it.

Answer 17

i think u should be able to to the rest (integral) from here....

Answer 18

yes, thank you so much!

Answer 19

yw.... :)

Answer 20

:)