Certain partial fractions can be done by the residual method, which avoids solving simultaneous equations.
(2-x)/(9-x^2) = (2-x) / ((3-x)(3+x))
set equal to
(2-x) / ((3-x)(3+x)) = A/(3-x) + B/(3+x)
Now here's the trick. Multiply both sides by (3-x) as x approaches 3.
(3-x)(2-x) / ((3-x)(3+x)) = A(3-x)/(3-x) + B(3-x)/(3+x)
(2-x)/(3+x) = A + B(3-x)/(3+x)
but remember, we chose x to approach 3, so substitute 3 for x in the above
(2-3)/(3+3) = A + B(0)/6
-1/6 = A
By similar reasoning,
(2-x)/(3-x) = B at x=-3
5/6 = B
So (2-x)/((3-x)(3+x)) = 1/6 * 1/(3-x) + 5/6 * 1/3+x)
and you can integrate from there. This trick works whenever no factor appears in the denominator of the fraction more than once. After practice, you can take the shortcut of just omitting one of the denominator factors in the original expression, and substituting for x, the value that would have made that factor 0.
This is what @como did above, give that guy best answer. My explanation is the engineering shortcut that we common use in control equations.