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).
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