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$$\int \frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}dx$$

I know I will be using partial fraction decomposition on this problem, at least it seems that way. so far, what I have is this:

$$\frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}=\frac{A}{x+3}{}+\frac{B}{(x+3)^2}+\frac{Cx+D}{x^2+3}$$

Multiplying by the LCD : $(x+3)^2(x^2+3)$

I am left with :

$$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2$$

By setting $x=-3:B=-4$

Now is where I am running into trouble. Now that I can substitute B into the original decomposition equation, There is no value of x that will leave only one variable to solve for. Please lend me a hand you guys(and girls). Thanks!

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    $\begingroup$ Expand the right hand side and compare the coefficients. $\endgroup$ – NECing Mar 17 '13 at 22:54
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There are many options:

  1. Differentiate and substitute $x=-3$.

  2. Substitute simple values such as $x=0$ and $x=1$ and $x=-1$ to get three equations in three unknowns.

  3. The method @Shu mentions in the comments.

  4. Substitute $x=\sqrt{-3}$ and $x=-\sqrt{-3}$.

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  • $\begingroup$ What happens with point 4 exactly? Do you mean: multiply both sides by $x^2+3$ first? $\endgroup$ – Julien Mar 17 '13 at 23:07
  • $\begingroup$ @julien, no multiplication, just substitution. On the right side, the terms with $A$ and $B$ disappear, you get two equations for $C$ and $D$. Maybe better is to realize $C$ and $D$ must be real, and just compare real and imaginary parts in one equation. Oh, I see, your are looking at the 2nd display in the question, I am looking at the 3rd. $\endgroup$ – Gerry Myerson Mar 17 '13 at 23:14
  • $\begingroup$ Ah yes, sure. I had forgotten the third equation...Sorry. $\endgroup$ – Julien Mar 17 '13 at 23:16
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I'll denote $f(x)$ this rational function. Here is an alternative to the expanding/equating strategy.

Limit trick: multiply both sides by $x$, compare the degrees, and let $x$ tend to $+\infty$: $$ \lim_{x\rightarrow+\infty}xf(x)=5=A+C. $$

Substitution trick: pick a small value in the domain of $f$. Here $0$ is perfect: $$ f(0)=\frac{-3}{27}=\frac{A}{3}+\frac{B}{3}+\frac{D}{3} $$ We need one more equation. Let's try $-2$, it's not too bad. $$ f(-2)=\frac{-21}{7}=A+B+\frac{D-2C}{7}. $$ Now, no miracle, there is a $4\times 4$ system to solve using also $B=-4$.

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$$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2=$$ $$=(A+C)x^3+(3A+B+6C)x^2+(3A+6D+9C)x+(9A+3B+9D)$$ equating the coefficients next to the same power of $x$ we get following system

$$A+C=5$$

$$3A+B+6C=19$$

$$3A+6D+9C=27$$

$$9A+3B+9D=-3$$

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\begin{align} 5x^3+19x^2+27x-3 &= A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2 &\text{Let $x = -3$}\\ -48&=12B \\ B &= -4 & \text{Go back and let $B = -4$}\\ \hline 5x^3+19x^2+27x-3 &= A(x+3)(x^2+3)-4(x^2+3)+(Cx+D)(x+3)^2 \\ 5x^3 + 23x^2 + 27x + 9 &= A(x+3)(x^2+3)+(Cx+D)(x+3)^2 &\text{Divide both sides by $(x+3)$}\\ \hline 5x^2 + 8x + 3 &= A(x^2+3)+(Cx+D)(x+3) &\text{Let $x=-3$} \\ 24 &= 12A \\ A &= 2 &\text{Go back and let A = 2} \\ \hline 5x^2 + 8x + 3 &= 2(x^2+3)+(Cx+D)(x+3) \\ 3x^2 + 8x - 3 &= (Cx+D)(x+3) &\text{Divide both sides by $(x+3)$} \\ 3x-1 &= Cx+D \\ C &= 3 \\ D &= -1 \end{align}

Computational note.

The quotient $(5x^3 + 23x^2 + 27x + 9) \div (x+3)$ and the value of that quotient at $x=-3$ can be accomplished with synthetic division.

\begin{array}{r|rrrrr} & 5 & 23 & 27 & 9 \\ -3 & 0 & -15 & -24 & -9 \\ \hline & 5 & 8 & 3 \\ -3 & 0 & -15 & 21\\ \hline & 5 & -7 & 24 \end{array}

Hence $(5x^3 + 23x^2 + 27x + 9) \div (x+3) = 5x^2+8x+3$ and the value of $5x^2+8x+3$ when $x=-3$ is $24$.

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