3/((x1)(x+2)) = A/(x1) + B/(x+2) for some numbers A and B.
Simplify both sides of the equation by multiplying by (x1)(x+2) [we can assume x not equal to 1 and 2].
We thus get
3 = A(x+2) + B(x1)
Plug in x = 2 to get 3 = 3B ==> B = 1.
Plug in x = 1 to get 3 = 3A ==> A = 1.
Thus INT[3/((x1)(x+2))]dx = INT[1/(x1)  1/(x+2)]dx =
INT[1/(x1)]dx + INT[1/(x+2)]dx.
Doing a Usubstitution on the first one we let U = x  1 ==> dU = dx.
Doing a Vsubstituion on the second one we let V = x + 2 ==> dV = dx.
Hence INT[1/(x1)]dx + INT[1/(x+2)]dx = INT[1/U]dU  INT[1/V]dV =
lnU  lnV since INT[1/x]dx = lnx.
Hence the Indefinite integral is
lnx1  lnx+2 + C for some C in the real numbers.
Now integrating from 2 to 3 should yield you ln(8/5).
1/31/2014

Adam S.
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