Evaluate the integral (Partial Fractions)
∫x^2+1/(x-3)(x-2)^2 dx

Why is it that when you split the function into A+B+C that you get A(x-2)+B(x-2)(x-3)+C(x-3). how come there isnt an A(x-2)^3

Question

Evaluate the integral (Partial Fractions)
∫x^2+1/(x-3)(x-2)^2 dx

Why is it that when you split the function into A+B+C  that you get A(x-2)+B(x-2)(x-3)+C(x-3). how come there isnt an A(x-2)^3

anonymous · Answer

There's no A(x-2)^3 because (x-2) is only raised to the 2nd power. You don't need the second (x-3) either.

$$\frac{ x^2+1 }{ (x-3)(x-2)^2 }=\frac{ A }{ x-3 }+\frac{ B }{ x-2 }+\frac{ C }{ (x-2)^2 }$$

anonymous · Answer

ok. How come there's C(x-3) and not C(x-2)(x-3). is it because there's a (x-2)^2 in the denominator?

anonymous · Answer

you mean when you multiply? You have to multiply by the least common denominator, so that's $(x-3)(x-2)^2$.

So (x -3) cancels on the A leaving you with $A(x-2)^2$
(x - 2) cancels on B, leaving $B(x-3)(x-2)$
(x - 2)² cancels on C leaving $C(x-3)$

anonymous · Answer

$$x^2+1=A(x-2)^2+B(x-3)(x-2)+C(x-3)$$

anonymous · Answer

ok its makes a lot of sense now. thanks.