206.5.64 integral by partial fractions

[karush]

$\textbf{206.5.64 integral by partial fractions} \\
\displaystyle
I_{64}=
\int\frac{9x^3-6x+4}{x^3-x^2} \, dx \\
\text{expand} \\
\displaystyle
\frac{9x^3-6x+4}{x^3-x^2}
= \frac{9(x^3-x^2)+9x^2+6x+4}{x^3-x^2}
= 9 + \frac{9x^2+6x+4}{x^2(x-1)} \\
\textbf{stuck!}$

 

[MarkFL]

The next step would be:

\(\displaystyle \frac{9x^2+6x+4}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}\)

 

[karush]

$\textbf{206.5.64 integral by partial fractions} \\
\displaystyle
I_{64}=
\int\frac{9x^3-6x+4}{x^3-x^2} \, dx =
2\ln\left(\left|x\right|\right)+9x+\dfrac{4}{x}+7\ln\left(\left|x-1\right|\right)\\
\textbf{expand} \\
\displaystyle
\frac{9x^3-6x+4}{x^3-x^2}
= \frac{9(x^3-x^2)+9x^2+6x+4}{x^3-x^2}
= 9 + \frac{9x^2+6x+4}{x^2(x-1)} \\
\textbf{sidework}\\
\displaystyle
\frac{9x^2+6x+4}{x^2(x-1)}
\displaystyle
=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}\\
\displaystyle 9x^2+6x+4=Ax(x-1)+B(x-1)+Cx^2=(A+C)x^2+(B-A)x+(-B) \\
4=(A+C-9)x^2+(B-A-6)x-B
$

 

[MarkFL]

We have;

\(\displaystyle \frac{9x^2+6x+4}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}\)

The next step is to multiply through by $x^2(x-1)$ to get:

\(\displaystyle 9x^2+6x+4=Ax(x-1)+B(x-1)+Cx^2=(A+C)x^2+(B-A)x+(-B)\)

Now, equate coefficients and solve the resulting system. :D