Partial Fractions with irreducible quadratics in denominator

Question

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?

Answer 1

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...

Answer 2

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