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1 Answer

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

Comments

Correction:   Result should be INT[1/U}dU - INT(1/V)dV  = ln|x-1| - ln|x+2| + C
 
So the definite integral from 2 to 3 is [ln(2) - ln(5)] - [ln(1) - ln(4)]  = ln(2/5 * 4/1) = ln(8/5).