I need some help at this exercise (partial fraction): $\int{\frac{4x + 1}{x^3(x+2)}} dx$
First of all I calculated the roots of the denominator. $ x_{1, 2, 3} = 0 \\ x_4 = -2 $
After that I assigned the roots to the partial fraction. $ \frac{A_1}{x}, \frac{A_2}{x^2}, \frac{A_3}{x^3}, \frac{B}{(x+2)} \\ \frac{4x + 1}{x^3(x+2)} = \frac{4x + 1}{x^3(x+2)} = \frac{A_1}{x} + \frac{A_2}{x^2} + \frac{A_3}{x^3} + \frac{B}{(x+2)} \\ 4x + 1 = A_1x^2(x+2) + A_2x(x+2) + A_3(x+2) + Bx^3 \\ 4x + 1 = A_1(x^3+2x^2) + A_2(x^2+2x) + A_3(x+2) + Bx^3 \\ $
Then I insert the root values. For $-2$, I get $B = \frac{7}{8}$. For $0$ I get $A_3 = \frac{1}{2}$. After that I insert two additional numbers, I choose $1$ and $-1$. I transformend the equation to $A_1$ and $A_2$ and get $A_1 = \frac{23}{20}$ and $A_2 = \frac{197}{120}$.
$ \frac{4x + 1}{x^3(x+2)} = \frac{\frac{23}{20}}{x} + \frac{\frac{197}{120}}{x^2} + \frac{\frac{1}{2}}{x^3} + \frac{\frac{7}{8}}{(x+2)} $
Could anyone tell me if it's correct? I belive it is wrong :), where is my misstake?