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anonymous
 one year ago
Evaluate the integral (Partial Fractions)
∫x^2+1/(x3)(x2)^2 dx
Why is it that when you split the function into A+B+C that you get A(x2)+B(x2)(x3)+C(x3). how come there isnt an A(x2)^3
anonymous
 one year ago
Evaluate the integral (Partial Fractions) ∫x^2+1/(x3)(x2)^2 dx Why is it that when you split the function into A+B+C that you get A(x2)+B(x2)(x3)+C(x3). how come there isnt an A(x2)^3

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There's no A(x2)^3 because (x2) is only raised to the 2nd power. You don't need the second (x3) either. \[\frac{ x^2+1 }{ (x3)(x2)^2 }=\frac{ A }{ x3 }+\frac{ B }{ x2 }+\frac{ C }{ (x2)^2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. How come there's C(x3) and not C(x2)(x3). is it because there's a (x2)^2 in the denominator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you mean when you multiply? You have to multiply by the least common denominator, so that's \((x3)(x2)^2\). So (x 3) cancels on the A leaving you with \(A(x2)^2\) (x  2) cancels on B, leaving \(B(x3)(x2)\) (x  2)² cancels on C leaving \(C(x3)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2+1=A(x2)^2+B(x3)(x2)+C(x3)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok its makes a lot of sense now. thanks.
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