MHB Is This Partial Fraction Decomposition Set Up Correct?

[karush]

$\tiny{206.8.5,42}\\$
$\textsf{partial fraction decomostion}\\$
\begin{align}
\displaystyle
&& I_{42}&=\int\frac{3x^2+x-18}{x^3+9x}\, dx& &(1)&\\
&& \frac{3x^2+x-18}{x^3+9x}
&=\frac{Ax+B}{x^2+9}
+\frac{C}{x}
& &(2)&
\end{align}

$\textit{just seeing if this is set up ok before finding values} $

 

[topsquark]

Re: partial fraction decomostion

$\tiny{206.8.5,42}\\$
$\textsf{partial fraction decomostion}\\$
\begin{align}
\displaystyle
&& I_{42}&=\int\frac{3x^2+x-18}{x^3+9x}\, dx& &(1)&\\
&& \frac{3x^2+x-18}{x^3+9x}
&=\frac{Ax+B}{x^2+9}
+\frac{C}{x^2+9}
+\frac{D}{x}
& &(2)&
\end{align}

$\textit{just seeing if this is set up ok before finding values} $

You are fine but you already accounted for the constant over x^2 + 9 when you included the B term. You don't need C.

-Dan

 

[karush]

Re: partial fraction decomostion

ok i rewrote OP
thot i could slip under the wire

I'll proceed..

 

[karush]

Re: partial fraction decomostion

$\tiny{206.8.5,42}\\$
$\textsf{partial fraction decomostion}\\$
\begin{align}
\displaystyle
&& I_{42}&=\int\frac{3x^2+x-18}{x^3+9x}\, dx& &(1)&\\
&& \frac{3x^2+x-18}{x^3+9x}
&=\frac{Ax+B}{x^2+9}-\frac{C}{x}& &(2)& \\
\\
&&3x^2+x-18&=(Ax+B)x-(x^2+9)C & &(3)& \\
&& &=Ax^2+Bx-Cx^2-9C & &(4)&\\
&x=0& -18&=-9C & &(5)&\\
&& 2&=C & &(6)&\\
&x=1& -14&=A+B-20 & &(7)&\\
&& 6&=A+B & &(8)&\\
&x=-1& && &(9)&\\
&& -14&=A-B-16 & &(10)&\\
&& 4&=A-B & &(11)& \\
&(8)+(11)& A&=5 \\
&\therefore (8)& B&=1
\end{align}

$\textit{just seeing if this is correct} $

 

[MarkFL]

Re: partial fraction decomostion

I would proceed only slight differently:

$$\frac{3x^2+x-18}{x^3+9x}=\frac{3x^2+x-18}{x(x^2+9)}=\frac{A}{x}+\frac{Bx+C}{x^2+9}$$

This implies

$$3x^2+x-18=A(x^2+9)+(Bx+C)x=(A+B)x^2+Cx+9A$$

Equating coefficients, we see:

$$9A=-18\implies A=-2$$

$$C=1$$

$$A+B=3\implies B=5$$

Hence:

$$\frac{3x^2+x-18}{x^3+9x}=-\frac{2}{x}+\frac{5x+1}{x^2+9}$$

This agrees with your result. :D

 

[karush]

Re: partial fraction decomostion

\begin{align}
\displaystyle
&& I_{42}&=\int\frac{3x^2+x-18}{x^3+9x}\, dx& &(1)&\\
\\
&& &=5\int\frac{x}{x^2+9}
+\int\frac{1}{x^2+9}
-2\int\frac{1}{x}& &(2)&\\
\\
&& &=\frac{5\ln\left({x^2+9}\right)}{2}
+\frac{\arctan(\frac{x}{3})}{3}
+2\ln\left({|x|}\right)+C & &(3)&
\end{align}

 

[MarkFL]

Re: partial fraction decomostion

What happened to the coefficient of $-2$ in front of the second log function? You could use a matrix I suppose, but this one is simple enough to make it unnecessary or beneficial, IMHO.

 

[karush]

Re: partial fraction decomostion

I corrected post 6

my excuse: tablet with small screen
my reason: none

 

[HallsofIvy]

Re: partial fraction decomostion

$\tiny{206.8.5,42}\\$
$\textsf{partial fraction decomostion}\\$
\begin{align}
\displaystyle
&& I_{42}&=\int\frac{3x^2+x-18}{x^3+9x}\, dx& &(1)&\\
&& \frac{3x^2+x-18}{x^3+9x}
&=\frac{Ax+B}{x^2+9}-\frac{C}{x}& &(2)& \\
\\
&&3x^2+x-18&=(Ax+B)x-(x^2+9)C & &(3)& \\
&& &=Ax^2+Bx-Cx^2-9C & &(4)&\\
&x=0& -18&=-9C & &(5)&\\
&& 2&=C & &(6)&\\
&x=1& -14&=A+B-20 & &(7)&\\
&& 6&=A+B & &(8)&\\
&x=-1& && &(9)&\\
&& -14&=A-B-16 & &(10)&\\

When x= -1, 3x^2+ x- 18= 3(1)+ (-1)- 18= 3- 19= -16

&& 4&=A-B & &(11)& \\
&(8)+(11)& A&=5 \\
&\therefore (8)& B&=1
\end{align}

$\textit{just seeing if this is correct} $