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I'm trying to solve this integral by partial fraction:

$$ \int \frac{2x-6} {(x-2)^2(x^2+4)} dx \ $$

i think i have to write the expression like

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

Then i don't know how i should resolve the partial fraction!

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    $\begingroup$ $-\frac{1}{4(x-2)^3}+\frac{5}{16(x-2)^2}-\frac{5}{64(x-2)}+\frac{5}{64(x+2)}$ $\endgroup$ – almagest Jun 22 '16 at 15:16
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    $\begingroup$ Your first step is correct only if in the first integral you have $x^2-4$ at the denominator. Is it a typo? $\endgroup$ – Emilio Novati Jun 22 '16 at 15:22
  • $\begingroup$ If so, I'll adjust my answer. Did you mean $x^2+4$? I suppose so, as the denominator seems to have been fully factorized this way. If it was in fact $x^2-4$, your next step is alright. $\endgroup$ – StackTD Jun 22 '16 at 15:28
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Careful!

$$ \int \frac{2x-6} {(x-2)^2(x^2+4)} dx \ $$

i think i have to write the expression like

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

You seem to have replaced $x^2+4$ by $(x-2)(x+2)$, but these are not the same! You might be confusing with $x^2\color{red}{-}4=(x-2)(x+2)$...

You can take out the (constant) factor $2$, but you don't have to. For the partial fraction decomposition, you're looking for numbers $A$, $B$, $C$ and $D$ such that: $$\frac{2x-6} {(x-2)^2(x^2+4)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4} $$ Once you have that, the integral splits into these three terms and these are all easy to integrate.

I can elaborate on the partial fraction decomposition; unless you can take it from here?


For the partial fraction decomposition:

$$\begin{array}{rl} \displaystyle \frac{2x-6} {(x-2)^2(x^2+4)} & \displaystyle = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{Cx+D}{x^2+4} \\[7pt] & \displaystyle = \frac{A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2}{(x-2)^2(x^2+4)} \end{array}$$

Now equating the numerators: $$2x-6 = A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2$$

Method 1

You can expand the RHS and group per power of $x$ and then identify the corresponding coefficients left and right; this gives you four linear equations in the 4 unknown variables.

$$2x-6 = (A+C)x^3+(-2A+B-4C+D)x^2+(4A+4C-4D)x+(-8A+4B+4D)$$

Which gives the system: $$\left\{\begin{array}{rcl} A+C & = & 0 \\ -2A+B-4C+D & = & 0 \\ 4A+4C-4D & =& 2 \\ -8A+4B+4D & = & -6 \end{array} \right. \quad \Rightarrow \quad \left\{\begin{array}{rcl} A & = & \tfrac{3}{8} \\ B & = & -\tfrac{1}{4} \\ C & =& -\tfrac{3}{8} \\ D & = & -\tfrac{1}{2} \end{array} \right.$$

Method 2

It is indeed also possible to simplify by choosing handy values for $x$ to substitute into $$2x-6 = A(x-2)(x^2+4)+B(x^2+4)+(Cx+D)(x-2)^2$$

For example:

  • Substitution of $x=2$ yields $-2=8B$ so $B = -\tfrac{1}{4}$.
  • Substitution of $x=0$ yields $-6=-8A+4B+4D$ so $8A-4D=5$.
  • (...)

If you know complex numbers, substitution of $x = 2i$ gives you $C$ and $D$ immediately, then $A$ follows from $8A-4D=5$.

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  • $\begingroup$ Ok, now i'm trying to solve A,B,Cx and D; If i put: x=2 i get B= -1/4 , then for x=0 i get -5=-8A-2D ,and i don't know what i've to do $\endgroup$ – Marco Jun 22 '16 at 15:41
  • $\begingroup$ Alright; note that you're just looking for the coefficient $C$, not $Cx$. Let me know if you have trouble with solving this. $\endgroup$ – StackTD Jun 22 '16 at 15:42
  • $\begingroup$ I know how to solve this when i only have A,B,C , but with 4 i can't follow after i solve the A $\endgroup$ – Marco Jun 22 '16 at 15:45
  • $\begingroup$ I edited my answer. $\endgroup$ – StackTD Jun 22 '16 at 16:02
  • $\begingroup$ Using the 2nd method, wich valor of x should i put in the 3rd substitution? , i think the idea is using (5+4D)/8 as A , then using 0 as x another time? $\endgroup$ – Marco Jun 22 '16 at 16:16

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