Is This Partial Fraction Decomposition Set Up Correct?

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Discussion Overview

The discussion revolves around the setup and execution of partial fraction decomposition for the integral of a rational function. Participants explore different approaches to expressing the integrand and share their calculations and reasoning.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents an initial setup for partial fraction decomposition, questioning its correctness.
  • Another participant suggests a modification to the decomposition, arguing that a constant term is unnecessary due to the presence of another term.
  • A participant rewrites the original problem and proceeds with calculations, indicating a desire to verify their work.
  • Further calculations are presented, showing the expansion of the integrand and the determination of coefficients.
  • Another participant proposes a different approach to the decomposition, leading to a consistent result with the previous calculations.
  • Subsequent posts address the integration process, with one participant noting a missing coefficient in the logarithmic term of the integral solution.
  • Corrections are made to earlier posts, with participants acknowledging the challenges of working on smaller screens.
  • One participant revisits the original setup and calculations, reaffirming their approach and seeking confirmation of correctness.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of certain terms in the partial fraction decomposition. While some calculations align, there is no clear consensus on the optimal approach or the correctness of all steps presented.

Contextual Notes

Some participants' calculations depend on specific assumptions about the decomposition structure, and there are unresolved steps in the mathematical reasoning that could affect the final results.

karush
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$\tiny{206.8.5,42}\\$
$\textsf{partial fraction decomostion}\\$
\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} $
 
Last edited:
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Re: partial fraction decomostion

karush said:
$\tiny{206.8.5,42}\\$
$\textsf{partial fraction decomostion}\\$
\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
 
Re: partial fraction decomostion

ok i rewrote OP
thot i could slip under the wire

I'll proceed..
 
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} $
 
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
 
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}
 
Last edited:
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.
 
Re: partial fraction decomostion

I corrected post 6

my excuse: tablet with small screen
my reason: none
 
Re: partial fraction decomostion

karush said:
$\tiny{206.8.5,42}\\$
$\textsf{partial fraction decomostion}\\$
\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} $
 

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