For what conditions the integral is a rational function.

Question

I have tried to solve a problem here and I've got an answer but I just don't know if it is the right one.

The problem is: For what conditions $$\int \frac{ax^2+bx+c}{x^3\left(x-1\right)^2}dx$$ represents a rational function.

First I have calculated the integral and the result is: $$\left(a+2b+3c\right)\ln\left|x\right|-\frac{2c+b}{x}-\left(a+2b+3c\right)\ln\left|x-1\right|-\frac{a+b+c}{x+1}-\frac{c}{2x^3}+K$$ where $K$ is a constant.

From here I have taken the conditions $$\left(a+2b+3c\:=\:0\right) \mbox{ and } ((b+2c != o) \mbox{ or } (a+b+c != 0))$$ which is the same as $$\left(a\:=-2b\:-3c\right) \mbox{ and } (b != -2c)$$

So, from my result, the condition above must be satisfied in order for that function to be a rational function.

P.S != means not equal to

Can anyone tell whether the solution is correct or not and if it is not then help me with it ?

Thank you very much!

Answer 1

The only condition you want is that the coefficient in front of the logarithms are zero, the rest is considered to be a rational function (also, a polynomial is considered to be a rational function). In your case, this means that $$ a+2b+3c= 0. $$ (The term $-(a+b+c)/(x+1)$ should be $(a+b+c)/(1-x)$, but that does not change the answer.)

Answer 2

Since $$\frac{1}{x^3\left(x-1\right)^2}=\frac{1}{(x-1)^2}-\frac{3}{x-1}+\frac{3}{x}+\frac{2}{x^2}+\frac{1}{x^3},$$ it follows that the integral of $(ax^2+bx+c)$ over $x^3\left(x-1\right)^2$ is rational iff, in the integrand function, the coefficient of $1/x$, that is $a+2b+3c$, and the coefficent of $1/(x-1)$, that is $-(a+2b+3c)$ are zero.

So the condition should be simply $a+2b+3c=0$.