$$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx$$ Here's what I have so far... $$\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)} = \frac{\mathrm A}{x+1}+\frac{\mathrm Bx+\mathrm C}{x^2+2x+5}\\$$ $$5x^2 + 9x + 16 = \mathrm A(x^2+2x+5) + (\mathrm Bx+\mathrm C)(x+1)=\\$$ $$\mathrm A(x^2+2x+5) + \mathrm B(x^2+x)+\mathrm C(x+1)=\\$$ $$(\mathrm A+\mathrm B)x^2 + (2\mathrm A + \mathrm B + \mathrm C)x + (5\mathrm A+\mathrm C)\\$$ $$\mathrm A=-3,\;\mathrm B=8,\;\mathrm C = 31$$ $$$$ $$\int\frac{5x^2+9x+16}{(x+1)(x^2+2x+5)}dx = \int\bigg(-\frac{3}{x+1}+\frac{8x+31}{x^2+2x+5}\bigg)dx\Rightarrow$$ $$\int-\frac{3}{x+1}dx +\int\frac{8}{x^2+2x+5}dx+\int\frac{31}{x^2+2x+5}dx $$

Hopefully I've got it correct until this point (if not, someone point it out please!). I can do the first integration by moving the -3 out and using $u=x+1$ to get $$-3 \ln(x+1)$$ but I'm stuck on the next two.

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    $\begingroup$ Is $2A+B+C=9$ for $A=-3,B=8,C=31$?. Doesn't seem right. $\endgroup$ – Thomas Andrews Sep 29 '14 at 2:25
  • $\begingroup$ I was just looking at that, actually. Checking it out now. $\endgroup$ – Jessica Sep 29 '14 at 2:25
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    $\begingroup$ Should be $A=3,B=2,C=1$. $\endgroup$ – Thomas Andrews Sep 29 '14 at 2:28
  • $\begingroup$ Is the last coefficient not 5A + C = 16? $\endgroup$ – Jessica Sep 29 '14 at 2:28
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    $\begingroup$ The second integral seems to be missing a factor of $x$ which might help simplify things. $\endgroup$ – Roger Burt Sep 29 '14 at 2:30

It should be $A=3,B=2,C=1$ and you get:

$$\int \frac{3}{x+1} dx + \int \frac{2x+2}{x^2+2x+5} dx -\int \frac{1}{x^2+2x+5}dx$$

The second is easy by using $u=x^2+2x+5$ and other answerers have covered the last.

The key is writing $\frac{2x+1}{x^2+2x+5} = \frac{2x+2}{x^2+2x+5} - \frac{1}{x^2+2x+5}$.

It's probably easier to just set $u=x+1$ at the start. Then you get:


Then write:

$$\frac{5u^2-u+12}{u(u^2+4)} = \frac{A}{u} + \frac{Bu+C}{u^2+4}$$ Giving $$A=3,B=2,C=-1$$, and the integrals:

$$\int\left(\frac{3}{u} + \frac{2u}{u^2+4} - \frac{1}{u^2+4}\right)\,du$$

  • $\begingroup$ could you explain how you get the numerators in the second and 3rd integrals? I thought it was just $\int\frac{a}{...}+\int\frac{bx+c}{...}$ which turned into $\int\frac{a}{...}+\int\frac{bx}{...}+\int\frac{c}{...}$? $\endgroup$ – Jessica Sep 29 '14 at 2:43
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    $\begingroup$ $2x+1$ can certainly be written as $2x+2-1$. It's easier to do the integral if you have $2x+2$ in the numerator, compared to $2x$, because the derivative of $x^2+2x+5$ is $2x+2$. $\endgroup$ – Thomas Andrews Sep 29 '14 at 2:46


Thomas Andrews says the values of $A,B,C$ are in error. I haven't checked those, but the technique outlined below still works if different numbers are involved.

end of PS

$$ \int\frac{8x+31}{x^2+2x+5}\,dx $$ First let $w=x^2+2x+5$ so that $dw=(2x+2)\,dx$. Then we have $$ \int\frac{8x+31}{x^2+2x+5}\,dx = 4\int\frac{2x+2}{x^2+2x+5}\,dx+ \int\frac{23\,dx}{x^2+2x+5} $$ Use the substitution to handle the first integral. Then $$ \overbrace{\int\frac{23\,dx}{x^2+2x+5} = \int\frac{23\,dx}{(x+1)^2+2^2}}^{\text{completing the square}} = \frac{23}2\int\frac{dx/2}{\left(\frac{x+1}{2}\right)^2+1} = \frac{23}2 \int\frac{du}{u^2+1} $$ and get an arctangent.

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    $\begingroup$ Not sure you know what PS means :) $\endgroup$ – Thomas Andrews Sep 29 '14 at 3:05
  • $\begingroup$ "postscript", i.e. something written later. $\endgroup$ – Michael Hardy Sep 29 '14 at 4:45
  • $\begingroup$ an additional remark at the end of a letter, after the signature and introduced by “P.S.” $\endgroup$ – Thomas Andrews Sep 29 '14 at 12:46
  • $\begingroup$ In a letter written on paper with a pen, it's necessarily at the end, but I don't think that's essential to the concept. This is a different format. What is essential is that it's added later. $\endgroup$ – Michael Hardy Sep 29 '14 at 14:03
  • $\begingroup$ Yeah, I was mostly kidding, but I have never seen PS used for a preamble note. The term "post script" is used more generally for notes added later, but "PS" has always been at the end of the text. $\endgroup$ – Thomas Andrews Sep 29 '14 at 14:09

(Just working on the last integral, but not checking the partial fractions.) Complete the square:

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

Now, substitute $x+1 = 2\tan(t)$. So, $dx = 2\sec^2 t\;dt$. Thus: $$\begin{align} \int \frac{dx}{(x+1)^2 + 4} &= \int\frac{2\sec^2(t)dt}{4(\tan^2t + 1)} \\ &= \frac{1}{2}\int\frac{\sec^2 t\; dt}{\sec^2t} \\ &= \frac{1}{2}\int 1\; dt\\ &= \cdots \end{align}$$

  • $\begingroup$ $\sin^2 t + 1\neq \cos^2 t$... $\endgroup$ – Thomas Andrews Sep 29 '14 at 2:30
  • $\begingroup$ @ThomasAndrews Should be better now. Thanks again. $\endgroup$ – apnorton Sep 29 '14 at 2:34
  • $\begingroup$ what would this look like back in terms of x? $\endgroup$ – Jessica Sep 29 '14 at 3:28
  • $\begingroup$ @Jessica Something along the lines of $\frac{1}{2}\arctan\left(\frac{x+1}{2}\right)$. (Just noticed another mistake--didn't cancel properly. doh.) $\endgroup$ – apnorton Sep 29 '14 at 3:30

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