# Partial Fractions with irreducible quadratics in denominator

I am not sure how to begin this question, as we have not really covered it in class:

$$\int \frac{3}{x^2 + 4x + 40}\, \mathrm dx$$

Correct me if I am wrong, but it doesn't seem that the denominator can be reduced further - how might I go about getting started on this?

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it doesn't seem that the denominator can be reduced further - how might I go about getting started on this?

That's usually the tipoff that you'll have to use an arctangent. Completing the square on the quadratic yields $(x+2)^2+36$; let $u=x+2$, and you should be able to see what to do next...

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AH ok. It has been a while since I have completed the square. Not sure where arctan figures in though... – Dylan Oct 29 '11 at 1:43
$$\int\frac{\mathrm du}{1+u^2}=\arctan\,u+C$$ is the applicable formula. – J. M. Oct 29 '11 at 1:47
Where did the 3 in the numerator go? And what about 36 from the previous completion of the square? – Dylan Oct 29 '11 at 4:50
1. $\int\frac{c\,\mathrm du}{1+u^2}=c\int\frac{\mathrm du}{1+u^2}$, right? 2. $\int\frac{\mathrm du}{a^2+u^2}$ is handled by letting $u=av$... – J. M. Oct 29 '11 at 6:54

$\begin{eqnarray*}\int\frac{3}{x^2 + 4x + 40}\, \mathrm dx & = & \frac{1}{12}\int\frac{1}{\left (\frac{x+2}{6}\right )^2 + 1}\, \mathrm dx\\ & & \left (\frac{x+2}{6}=t,\mathrm dx=6\mathrm dt\right )\\ & = & \frac{1}{2}\int\frac{1}{t^2 + 1}\, \mathrm dt\\ & & \left (t=\tan{u},\mathrm dt=\sec^2{u}\mathrm du\right )\\ & = & \frac{1}{2}\int\frac{\sec^2{u}}{\tan^2{u} + 1}\, \mathrm du=\frac{1}{2}\int\frac{\sec^2{u}}{\sec^2{u}}\, \mathrm du=\frac{1}{2}u+\mathrm{Const}\\ &=&\frac{1}{2}\arctan{\left ( \frac{x+2}{6} \right)}+\mathrm{Const} \end{eqnarray*}$

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By this stage, $\int\frac{dt}{1+t^2}$ is usually one of the "standard" known antiderivatives, so the $t=\tan u$ substitution may not be necessary. – André Nicolas Oct 29 '11 at 2:38
not sure where $(\frac{x+2}{6}\)^2$ comes from. When I completed the square, I got (x+2)^2 + 36 – Dylan Oct 29 '11 at 4:49
@Dylan: $\frac{3}{36\times\left \{\left (\frac{x+2}{6}\right )^2 + 1 \right \}}$... – Xiang Oct 29 '11 at 5:58
hmm, i am afraid that still doesn't clarify it for me. forgive me - i am new to integrals. i will be sure to consult the prof for the next class. – Dylan Oct 29 '11 at 8:06