<s>[INST] $\int{\frac{x^3}{(x-1)(x-2)(x-3)}}dx$ Question:
$$Find\int{\frac{x^3}{(x-1)(x-2)(x-3)}}dx$$
What I did:
resolved into partial fraction;
$$A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) = x^3$$
$$A = \frac{1}{2},\hspace{0.2cm} B = -8\hspace{0.2cm}, C= \frac{27}{2}$$
$$\frac{1}{2}\int\frac{1}{x-1}dx \hspace{0.2cm}-8\int\frac{1}{x-2} + \hspace{0.2cm}\frac{27}{2}\int\frac{1}{x-3}$$
$$= \frac{1}{2}ln(x-1)\hspace{0.2cm} -8ln(x-2)\hspace{0.2cm}+ \frac{27}{2}ln(x-3) + c$$
But it was done differently in the textbook and other answers I see, they did long division and changed the form to $Quotient + \frac{Remainder}{Divisor}$:
$$\int{\frac{x^3}{(x-1)(x-2)(x-3)}}dx$$
$$= \int\left(1 + \frac{6x^2 - 11x + 6}{(x-1)(x-2)(x-3)}\right)dx$$
$$A = \frac{1}{2},\hspace{0.2cm} B = -8\hspace{0.2cm}, C= \frac{27}{2}$$
$$\int dx \hspace{0.2cm}+\frac{1}{2}\int\frac{1}{x-1}dx \hspace{0.2cm}-8\int\frac{1}{x-2} + \hspace{0.2cm}\frac{27}{2}\int\frac{1}{x-3}$$
$$=x + \hspace{0.2cm} \frac{1}{2}ln(x-1)\hspace{0.2cm} -8ln(x-2)\hspace{0.2cm}+ \frac{27}{2}ln(x-3)+ c$$
The answers are different, my question is, what did I do wrong in the first solution? To the best of my knowledge I think I did the partial fraction correctly.
 [/INST] The equation$$A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)=x^3$$has no solutions, since the LHS is a quadratic polynomial, whereas the RHS is a cubic one. And actually$$\frac12(x-2)(x-3)-8(x-1)(x-3)+\frac{27}2(x-1)(x-2)=6x^2-11x+6.$$
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