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Find the integral using partial factions.

$$\int\frac{(2x^2+5x+3)}{(x-1)(x^2+4)}\,dx$$

So do I do...$$\frac{2x^2+5x+3}{(x-1)(x^2+4)}=\frac{A}{x-1} + \frac{Bx+C}{x^2+4},$$

then get \begin{align*} 2x^2+5x+3 &= A(x^2+4)+(Bx+C)(x-1) \\ 2x^2+5x+3 &= Ax^2+4A+Bx^2-Bx+Cx-C? \end{align*}

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    $\begingroup$ Welcome to MSE! What have you tried and where are you stuck? Regards $\endgroup$ – Amzoti Nov 13 '13 at 2:15
  • $\begingroup$ I'm stuck at multiplying by (x-1)(x^2-4) through A/(x-1) + Bx+/(x^2+4)... I get Ax^2-4A+Bx^2-Bx+Cx-C? Is that right? $\endgroup$ – Itzel Nov 13 '13 at 2:52
  • $\begingroup$ @Itzel You should multiply by $(x - 1)(x^2 + 4)$, not $(x - 1)(x^2 - 4)$. $\endgroup$ – glebovg Nov 13 '13 at 5:00
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Note that we can simply add zero: $2x^2 = 2(x^2 + 4) - 8 = 2x^2 + 8 - 8$. Then $$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2(x^2 + 4) - 8 + 5x + 3}{(x - 1)(x^2 + 4)} = \frac{2}{x - 1} + \frac{5x - 5}{(x - 1)(x^2 + 4)}.$$ Do you think you can integrate $$\frac{2x^2 + 5x + 3}{(x - 1)(x^2 + 4)}$$ now?

Hint: $$\frac{5x - 5}{(x - 1)(x^2 + 4)} = \frac{5(x - 1)}{(x - 1)(x^2 + 4)} = \frac{5}{x^2 + 4}$$ and $$\int \frac{1}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C, \quad a \ne 0.$$

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  • $\begingroup$ Where did you get 5x-5? $\endgroup$ – Itzel Nov 13 '13 at 2:41
  • $\begingroup$ Note: $5x - 8 + 3 = 5x - 5$ and $(A + B)/C = A/C + B/C$. $\endgroup$ – glebovg Nov 13 '13 at 2:43
  • $\begingroup$ To expect the typical student to come up with such "notes" is ludicrous. Maybe you can do polynomial division/partial fractions in your head, but most students cannot... $\endgroup$ – The Chaz 2.0 Nov 13 '13 at 2:49
  • $\begingroup$ What? There is no polynomial division here, I simply added zero: $2x^2 = 2(x^2 + 4) - 8 = 2x^2 + 8 - 8$. This is Algebra I. Please read before you downvote my answer for no reason. $\endgroup$ – glebovg Nov 13 '13 at 2:56
  • $\begingroup$ Sorry I didn't mean to down vote your answer. I don't even think I can or it wasn't me. $\endgroup$ – Itzel Nov 13 '13 at 3:00
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It seems that you are stuck at the actual partial fraction decomposition, rather than at the integration. So let's pick up at $$2x^2+5x+3= A(x^2+4) + (Bx+C)(x-1)$$

While this must hold true for all values of $x$, certain values of $x$ will lead us to the values of $A,B,C$ more quickly...

$x = 1$ is a natural choice, as that will make the $(Bx+C)$ go away. We then have $$2(1)^2 + 3(1) + 5 = A((1)^2 + 4) + 0$$ $$2 + 3 + 5 = 5A$$ $$A=2$$

Then you could use complex x-values ($\pm 2i$ for example) and equate real and imaginary parts, but you could just pick a few other small (real) x-values for ease of computation.

So $x=2$ gives us $21 = 16 + 2B + C$, or $$2B +C = 5$$

Then $x = 3$ gives us $36 = 26 + (3B+C)(2)$, or $$3B+C = 5$$

Solving these last two equations simultaneously gives $B = 0, C = 5$.

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Hint: Write out the fraction given as $\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$ and equate the numerator to $2x^2+5x+3$. And the ensuing integral should be easy.

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  • $\begingroup$ Hi can you help me out? $\endgroup$ – Itzel Nov 13 '13 at 3:05

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